Question 14 Marks
Differentiate the following from first principle$\sin\sqrt{2\text{x}}$
AnswerWe have,$\text{f}(\text{x})=\sin\sqrt{2\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sin\sqrt{2\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\Big(\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{2}\Big)\cos\Big(\frac{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Bigg(\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{2}\Bigg)(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}})}{\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})}{2}\Bigg)(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}})\text{h}}\cos\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}})}{2}\Bigg)$$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})}{2}\Bigg)}{\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})}{2}\Bigg)}\lim_\limits{\text{h}\rightarrow0}\frac{{2(\text{x}+\text{h})}-{2\text{x}}}{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})\text{h}}\lim_\limits{\text{h}\rightarrow0}\cos\Bigg(\frac{\sqrt{2(\text{x}+\text{h})+\text{2x}}}{2}\Bigg)$
$=1\times\frac{2}{2\sqrt{2\text{x}}}\cos(\sqrt{2\text{x}})$
$=\frac{\cos(\sqrt{2\text{x}})}{\sqrt{2\text{x}}}$
View full question & answer→Question 24 Marks
Differentiate the following from first principle$3^{\text{x}^2}$
Answer$\text{f}(\text{x})=3^{\text{x}^2}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$\frac{\text{d}}{\text{dx}}\big(3^{\text{x}^2}\big)=\lim_\limits{\text{h}\rightarrow0}\frac{3^{(\text{x}+\text{h})^2}-3^{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{3^{\text{x}^2+\text{2xh}+\text{h}^2}-3^{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{3^{\text{x}^2}(3^{\text{x}^2+\text{2xh}+\text{h}^2-\text{x}^2}-1)}{\text{h}}\times\frac{(\text{h}+\text{2x})}{(\text{h}+\text{2h})}$
$=3^{\text{x}^2}\lim_\limits{\text{h}\rightarrow0}\frac{3^{\text{h}(\text{h}+\text{2x})}-1}{\text{h}(\text{h}+\text{2x})}\lim_\limits{\text{h}\rightarrow0}(\text{h}+\text{2x})$
$=3^{\text{x}^2}\log3(2\text{x})$
$=2\text{x}3^{\text{x}^2}\log3$
View full question & answer→Question 34 Marks
Differentiate the following from the first principle$\text{x}^2\sin\text{x}$
AnswerWe have, $\text{f}(\text{x})=\text{x}^2\sin\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})^2\sin(\text{x}+\text{h})-\text{x}^2\sin\text{x}}{\text{h}}$
$\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}^2+\text{h}^2+\text{2hx})(\sin\text{x}.\cos\text{x}+\cos\text{x}\sin\text{h})-\text{x}^2\sin\text{x}}{\text{h}}$ $[\because\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}^2\sin\text{x}(\cos\text{h}-1)}{\text{h}}+\frac{\text{h}(\text{h}+\text{2x})\sin\text{x}.\cos\text{h}}{\text{h}}+(\text{x}+\text{h})^2\cos\text{x}\frac{\sin\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}-\text{x}^2\sin\text{x}\times\frac{2\sin^2\frac{\text{h}}{2}}{\Big(\frac{\text{h}}{2}\Big)^2}\times\frac{\text{h}^2}{4}+(\text{h}+\text{2x})\sin\text{x}.\cos\text{2h}+(\text{x}+\text{h})^2\cos\text{x}$
$=0+(2\text{x}\sin\text{x}+\text{x}^2\cos\text{x})$
$=\text{2x}\sin\text{x}+\text{x}^2\cos\text{x}$
View full question & answer→Question 44 Marks
Differentiate the following from the first principle$\sqrt{\sin\text{2x}}$
AnswerWe have,$\text{f}(\text{x})=\sqrt{\sin\text{2x}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sqrt{\sin\text{2x}}}{\text{h}}$
Multiplying numerator and denominator by $\Big(\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin\text{2x}}\Big)$
$\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sqrt{\sin\text{2x}}}{\text{h}}\times\frac{\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin\text{2x}}}{\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin2\text{x}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h})-\sin\text{2x}}{\Big(\sqrt{\sin(\text{2x}+\text{2h})}+\sqrt{\sin2\text{x}}\Big)}$ $\Big[\sin\text{c}-\sin\text{d}=2\cos\frac{\text{c}+\text{d}}{2}\sin\frac{\text{c}-\text{d}}{2}\Big]$
$\lim_\limits{\text{h}\rightarrow0}\frac{2\cos(\text{2x}+\text{h})\times\sin\text{h}}{\text{h}}\times\frac{1}{\sqrt{\sin(\text{2x}+\text{2h})}+\sqrt{\sin\text{2x}}}$
$=\frac{2\cos\text{2x}}{2\sqrt{\sin\text{2x}}}$
$=\frac{\cos\text{2x}}{\sqrt{\sin\text{2x}}}$
View full question & answer→Question 54 Marks
For the function $\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+\dots+\frac{\text{x}^2}{2}+\text{x}+1.$Prove that $\text{f}'(1)=100\text{f}'(0).$
AnswerWe have,$\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+\dots+\frac{\text{x}^2}{2}+\text{x}+1.$
Differentiating with respect to $\text{x},$ We get
$\text{f}'(\text{x})=\text{x}^{99}+\text{x}^{98}+\dots+\text{x}+1+0\dots(\text{i})$
From (i)
f'(1) = 1 + 1 + ...(100 times)
= 100
Again,
f'(0) = 0 + 0 + ... + 1
= 1
Now,
f'(1) = 100 = 100 $\times$ 1 = 100 $\times$ f'(0)
Hence'
f'(1) = 100f'(0)
View full question & answer→Question 64 Marks
Differentiate the following from first principle$\sqrt{\tan\text{x}}$
AnswerWe have,$\text{f}(\text{x})=\sqrt{\tan\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\tan(\text{x}+\text{h})}-\sqrt{\tan\text{x}}}{\text{h}}$
Multiplying Numerator and Denominator by $\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan(\text{x}+\text{h})-\tan{\text{x}}}{\text{h}\big(\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big)}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}+\text{h}-\text{x})}{\text{h}.\cos(\text{x}+\text{h})\cos\text{x}\big[\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big]}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{1}{\cos(\text{x}+\text{h})\cos\text{x}\big[\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big]}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{1}{\cos^2\text{x}.2\sqrt{\tan\text{x}}}\ \Bigg[\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Bigg]$
$=\frac{1}{2}\frac{\sec^2\text{x}}{\sqrt{\tan\text{x}}}\ \Bigg[\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Bigg]$
View full question & answer→Question 74 Marks
Differentiate the following from first principle$\tan\text{2x}$
Answer$\text{f}(\text{x})=\tan\text{2x}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan2(\text{x}+\text{h})-\tan2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}-\text{2x})}{\text{h}.\cos(\text{2x}+\text{2h})\cos\text{2x}}\ \Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin\text{A}-\text{B}}{\cos\text{A}.\cos{\text{B}}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{\text{h}.\cos(\text{2x}+\text{2h})\cos2\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin2\text{h}}{2\text{h}}\Big)\times\frac{1\times2}{\cos(\text{2h}+\text{2x})\cos2\text{x}}$
$=\frac{2}{\cos2\text{x}.\cos2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2\text{h}}=1\Big]$
$=2\sec^22\text{x}\ \Big[\because\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Big]$
View full question & answer→Question 84 Marks
Differentiate the following from first principle$\tan\text{2x}$
Answer$\text{f}(\text{x})=\tan\text{2x}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan2(\text{x}+\text{h})-\tan2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}-\text{2x})}{\text{h}.\cos(\text{2x}+\text{2h})\cos\text{2x}}\ \Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin\text{A}-\text{B}}{\cos\text{A}.\cos{\text{B}}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{\text{h}.\cos(\text{2x}+\text{2h})\cos2\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin2\text{h}}{2\text{h}}\Big)\times\frac{1\times2}{\cos(\text{2h}+\text{2x})\cos2\text{x}}$
$=\frac{2}{\cos2\text{x}.\cos2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2\text{h}}=1\Big]$
$=2\sec^22\text{x}\ \Big[\because\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Big]$
View full question & answer→Question 94 Marks
Differentiate the following from first principle$\tan\text{x}^2$
AnswerWe have,$\text{f}(\text{x})=\tan\text{x}^2$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan{(\text{x}+\text{h})^2}-\tan{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2}{\cos(\text{x}+\text{h})}-\frac{\sin\text{x}^2}{\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2\cos\text{x}^2-\cos(\text{x}+\text{h})^2\sin\text{x}^2}{\cos(\text{x}+\text{h})^2\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big((\text{x}+\text{h})^2-\text{x}^2\Big)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}^2+\text{h}^2+\text{2xh}-\text{x}^2)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h}^2+\text{2xh})}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2 }$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{(\text{h}+\text{2x})}{\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=1.\frac{\text{2x}}{\cos^2(\text{x})^2}$
$=\text{2x}\sec^2\text{x}^2$
View full question & answer→Question 104 Marks
Differentiate the following from the first principle$\cos\Big(\text{x}-\frac{\pi}{8}\Big)$
AnswerLet $\text{f}(\text{x})=\cos\Big(\text{x}-\frac{\pi}{8}\Big).$ Then, $\text{f}(\text{x}+\text{h})=\cos\Big(\text{x}+\text{h}-\frac{\pi}{8}\Big)$$\therefore\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\cos\Big(\text{x}+\text{h}-\frac{\pi}{8}\Big)-\cos\Big(\text{x}-\frac{\pi}{8}\Big)}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Bigg[\frac{\big(\text{x}+\text{h}-\frac{\pi}{8}\big)+\big(\text{x}-\frac{\pi}{8}\big)}{2}\Bigg]\sin\Bigg[\frac{\big(\text{x}+\text{h}-\frac{\pi}{8}\big)-\big(\text{x}-\frac{\pi}{8}\big)}{2}\Bigg]}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Bigg[\frac{\text{2x}+\text{h}-\frac{2\pi}{8}}{2}\Bigg]\sin\Big[\frac{\text{h}}{2}\Big]}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}-\sin\Bigg[\frac{2\text{x}+\text{h}-\frac{2\pi}{8}}{2}\Bigg]\times\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big[\frac{\text{h}}{2}\Big]}{\frac{\text{h}}{2}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=-\sin\Bigg[\frac{\text{2x}+0-\frac{2\pi}{8}}{2}\Bigg]\times1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=-\sin\Big(\text{x}-\frac{\pi}{8}\Big)$
View full question & answer→Question 114 Marks
Differentiate the following from first principle$\tan\text{x}^2$
AnswerWe have,$\text{f}(\text{x})=\tan\text{x}^2$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan{(\text{x}+\text{h})^2}-\tan{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2}{\cos(\text{x}+\text{h})}-\frac{\sin\text{x}^2}{\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2\cos\text{x}^2-\cos(\text{x}+\text{h})^2\sin\text{x}^2}{\cos(\text{x}+\text{h})^2\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big((\text{x}+\text{h})^2-\text{x}^2\Big)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}^2+\text{h}^2+\text{2xh}-\text{x}^2)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h}^2+\text{2xh})}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2 }$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{(\text{h}+\text{2x})}{\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=1.\frac{\text{2x}}{\cos^2(\text{x})^2}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$
$=\text{2x}\sec^2\text{x}^2$
View full question & answer→Question 124 Marks
Differentiate the following from first principle$\cos\sqrt{\text{x}}$
AnswerWe have,$\text{f}(\text{x})=\cos\sqrt{\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\cos(\text{x}+\text{h})}-\cos\sqrt{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Big(\frac{\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}}}{2}\Big)\sin\Big(\frac{\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}}}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Bigg(\frac{\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}}}{2}\Bigg)(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})}{\Bigg(\frac{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{2}\Bigg)(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})\text{h}}\sin\Bigg(\frac{(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})}{2}\Bigg)$$=\lim_\limits{\text{h}\rightarrow0}\frac{-\sin\Bigg(\frac{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{2}\Bigg)}{\Bigg(\frac{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{2}\Bigg)}\times\frac{{\text{x}+\text{h}}-{\text{x}}}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})\text{h}}\times\sin\Bigg(\frac{\sqrt{(\text{x}+\text{h})+\text{x}}}{2}\Bigg)$
$=\lim_\limits{\text{h}\rightarrow0}-1\frac{\text{h}}{\text{h}(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})}\times\sin\Bigg(\frac{\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}}{2}\Bigg)$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-1}{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}\sin\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{2}$
$=\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
View full question & answer→Question 134 Marks
Differentiate the following from first principle$\tan^2\text{x} $
Answer$\text{f}(\text{x})=\tan^2\text{x}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan^2(\text{x}+\text{h})-\tan^2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\tan(\text{x}+\text{h})+\tan{\text{x}}\big\}\big\{\tan(\text{x}+\text{h})-\tan{\text{x}}\big\}}{\text{h}}$
$\big[\because\tan^2\text{A}-\tan^2\text{B}=(\tan\text{A}+\tan\text{B})(\tan\text{A}-\tan\text{B})\big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h}+\text{x})}{\cos(\text{x}+\text{h})\cos\text{x}}\times\frac{\sin(\text{x}+\text{h}-\text{x})}{\cos(\text{x}+\text{h})\cos\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{h})}{\text{h}.\cos(\text{x}+\text{h})\cos\text{x}}\times\frac{\sin\text{h}}{\cos(\text{x}+\text{h})\cos\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{\sin2\text{x}}{\cos^2\text{x}.\cos^2(\text{x}+\text{h})}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{x}}{\cos^2\text{x}.\cos^2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\text{x}.\cos\text{x}}{\cos^2\text{x}}\times\frac{1}{\cos^2\text{x}}\ [\sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\lim_\limits{\text{h}\rightarrow0}2\tan\text{x}.\sec^2\text{x}$
View full question & answer→Question 144 Marks
Differentiate the following from first principle:$-\text{x}$
AnswerLet $\text{f}\text{(x)}=-\text{x.}$ Then, $\text{f}(\text{x}+\text{h})=(-\text{x}+\text{h})$$\therefore\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-(\text{x}+\text{h})+(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}-1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=-1$
View full question & answer→Question 154 Marks
Differentiate the following from first principle$\tan\sqrt{\text{x}}$
AnswerWe have,$\text{f}(\text{x})=\tan\sqrt{\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan\sqrt{(\text{x}+\text{h})}-\tan\sqrt{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}.\cos\sqrt{\text{x}+\text{h}}\cos\sqrt{\text{x}}} \ \Bigg[\because\tan\text{A}-\tan\text{B}=\frac{\sin(\text{A}-\text{B})}{\cos\text{A}.\cos\text{B}}\Bigg]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\text{x}+\text{h}-\text{x})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}\times\frac{1}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=1\times\frac{1}{2\sqrt{\text{x}}\cos\sqrt{\text{x}}\cos\sqrt{\text{x}+\text{h}}}\ \Bigg[\because\lim_\limits{\text{h}\rightarrow0}=\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}=1\Bigg]$
$=\frac{1}{2\sqrt{\text{x}}\cos^2\text{x}}$
$=\frac{\sec^2\text{x}}{2\sqrt{\text{x}}}$
View full question & answer→Question 164 Marks
Differentiate in two ways, using product rule and otherwise, the function $(1+2\tan\text{x})(5+4\cos\text{x}).$ verify that the answers are the same.
AnswerLet $\text{u}=1+2\tan\text{x};\text{v}=5+4\cos\text{x}$Then, $\text{u}'=2\sec^2\text{x};\text{v}'=-4\sin\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(1+2\tan\text{x})(5+4\cos\text{x})]\\=(1+2\tan\text{x})(-4\sin\text{x})+(5+4\cos\text{x})(2\sec^2\text{x})$
$=-4\sin\text{x}-8\tan\text{x}\sin\text{x}+10\sec^2\text{x}+8\sec\text{x}$
$=-4\sin\text{x}+10\sec^2\text{x}+\Big(\frac{8}{\cos\text{x}}-\frac{8\sin^2\text{x}}{\cos\text{x}}\Big)$
$=-4\sin\text{x}+10\sec^2\text{x}+8\Big(\frac{1-\sin\text{x}}{\cos\text{x}}\Big)$
$=-4\sin\text{x}+10\sec^2\text{x}+8\Big(\frac{\cos^2\text{x}}{\cos\text{x}}\Big)$
$=-4\sin\text{x}+10\sec^2\text{x}+8\cos\text{x}$
Alternate Answer
$(1+2\tan\text{x})(5+4\cos\text{x})=5+4\cos\text{x}+10\tan\text{x}+8\sin\text{x}$
Now we have,
$\frac{\text{d}}{\text{dx}}[(1+2\tan\text{x})(5+4\cos\text{x})]\\=\frac{\text{d}}{\text{dx}}(5+4\cos\text{x}+10\tan\text{x}+8\sin\text{x})$
$=-4\sin\text{x}+10\sec^2\text{x}+8\cos\text{x}$
View full question & answer→Question 174 Marks
Differentiate the following from the first principle$\sin\text{x}+\cos\text{x}$
AnswerWe have, $\text{f}(\text{x})=\sin\text{x}+\cos\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\sin(\text{x}+\text{h})+\cos(\text{x}+\text{h})\big\}-\sin\text{x}+\cos\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\sin(\text{x}+\text{h})+\cos(\text{x}+\text{h})-\sin\text{x}-\cos\text{x}\big\}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\sin(\text{x}+\text{h})-\sin\text{x}\big\}+\big\{\cos(\text{x}+\text{h})-\cos\text{x}\big\}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big\{2\sin\frac{(\text{x}+\text{h}-\text{x})}{2}\cos\frac{(\text{x}+\text{h}+\text{x})}{2}\Big\}+\Big\{-2\sin\frac{\text{x}+\text{h}+\text{x}}{2}\sin\frac{\text{x}+\text{h}-\text{x}}{2}\Big\}}{\text{h}}$
$\begin{bmatrix}\because\sin\text{A}-\sin\text{B}=2\sin\frac{\text{A}-\text{B}}{2}\cos\frac{\text{A}+\text{B}}{2}\\\cos\text{A}-\cos\text{B}=2\sin\frac{\text{A}+\text{B}}{2}\cos\frac{\text{A}-\text{B}}{2}{}\end{bmatrix}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}.\cos\frac{\text{2x}+\text{h}}{2}-2\sin\Big(\text{x}+\frac{\text{h}}{2}\Big)\sin\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\Big\{\cos\frac{\text{x}+\text{h}}{2}-\sin\Big(\text{x}+\frac{\text{h}}{2}\Big)\Big\}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$
$=\cos\text{x}-\sin\text{x}$
View full question & answer→Question 184 Marks
Differentiate the following from first principle:$\frac{\text{x}+1}{\text{x}+2}$
AnswerWe have,$\text{f(x)}=\frac{\text{x}+1}{\text{x}+2}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{(x+h)+1}}{\text{(x+h)+2}}-\frac{\text{x}+1}{\text{x}+2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{(x+2)}\text{(x+h+1)}-\text{(x+1)(x+h+2)}}{\text{(x+h+2)}\text{(x+2)h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{x}^2+2\text{x+xh+2h+2+x}\big)-\big(\text{x}^2+\text{xh+2x+x+h+2}\big)}{\text{(x+h+2)(x+2)h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{(x+h+2)(x+2)h}}$
$=\frac{1}{(\text{x}+2)^2}$
View full question & answer→Question 194 Marks
Differentiate the following from the first principle$\text{x}\sin\text{x}$
AnswerWe have,$\text{f}(\text{x})=\text{x}\sin\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})\sin(\text{x}+\text{h})-\text{x}\sin\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}(\sin(\text{x}+\text{h})-\sin\text{x})}{\text{h}}+\sin(\text{x}+\text{h})\ $ $\Big[\sin\text{c}-\sin\text{d}=2\cos\frac{\text{c}+\text{d}}{2}\sin\frac{\text{c}-\text{d}}{2}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\times2\cos\Big(\text{x}+\frac{\text{h}}{2}\Big)\sin\frac{\text{h}}{2}}{\text{h}}+\sin(\text{x}+\text{h})\ $ $\Big[\because\lim_\limits{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\text{2x}\times\cos\text{x}\times\frac{1}{2}+\sin\text{x}$
$=\text{x}\times\cos\text{x}+\sin\text{x}$
$=\sin\text{x}+\text{x}\cos\text{x}$
View full question & answer→Question 204 Marks
Differentiate the following from the first principle$\sin(2\text{x}-3)$
AnswerWe have,$\text{f}(\text{x})=\sin(\text{2x}-3)$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\big\{2(\text{x}+\text{h})-3\big\}-\sin(\text{2x}-3)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\cos\frac{(\text{2x}+\text{2h}-3+\text{2x}-3)}{2}\times\sin\frac{(\text{2x}+\text{2h}-3-2-\text{x}+3)}{2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}2\cos(\text{2x}-3+\text{h}).\frac{\sin\text{x}}{\text{x}}$
$=2\cos(\text{2x}-3)$
View full question & answer→Question 214 Marks
Differentiate the following from the first principle$\text{x}^2\text{e}{^\text{x}}$
AnswerWe have, $\text{f}(\text{x})=\text{x}^2\text{e}^\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}^2\text{e}^\text{x}\text{e}^\text{h}+\text{h}^2\text{e}^\text{x}\text{e}^\text{h}+2\text{xhe}^\text{x}\text{e}^\text{h}-\text{x}^2\text{e}^\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\text{x}^2\text{e}^\text{x}\frac{(\text{e}^\text{h}-1)}{\text{h}}+\text{e}^\text{x}\text{e}^\text{h}\frac{(\text{h}^2+\text{2xh})}{\text{h}}\ \Big[\because\frac{\text{e}^\text{h}-1}{\text{h}}-1\Big]$
$\therefore\text{x}^2\text{e}^\text{x}+\text{e}^\text{x}(0+\text{2x})$
$=\text{x}^2\text{e}^\text{x}+\text{2xe}^\text{x}$
$=\text{e}^\text{x}(\text{x}^2+\text{2x})$
View full question & answer→Question 224 Marks
Differentiate the following functions with respect to x:$\frac{1}{\text{ax}^2+\text{bx}+\text{c}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{ax}^2+\text{bx}+\text{c}}\Big)$
$=\frac{(\text{ax}^2+\text{bx}+\text{c})\frac{\text{d}}{\text{dx}}(1)-(1)\frac{\text{d}}{\text{dx}}(\text{ax}^2+\text{bx}+\text{c})}{(\text{ax}^2+\text{bx}+\text{c})^2}$
$=\frac{(\text{ax}^2+\text{bx}+\text{c})(0)-(1)(\text{2ax}+\text{b})}{(\text{ax}^2+\text{bx}+\text{c})^2}$
$=\frac{-(\text{2ax}+\text{b})}{(\text{ax}^2+\text{bx}+\text{c})^2}$
View full question & answer→Question 234 Marks
Differentiate the following from first principle:$(\text{x}+2)^3$
AnswerWe have,$\text{f(x)}=\text{(x+2)}^3$
$\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\big(\text{x+h}\big)-\text{f}\big(\text{x}\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{x}+\text{h}+2\big)^3-\big(\text{x+2}\big)^3}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg\{\big(\text{x+2}\big)+\text{h}\bigg\}^3-\big(\text{x+h}\big)^3}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{x+2}\big)^3+\text{h}^3+3\text{h}\big(\text{x+2}\big)^2+3\big(\text{x+2}\big)\text{h}^2-\big(\text{x+2}\big)^3}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}3\big(\text{x+2}\big)^2+3\big(\text{x+2}\big)\text{h+h}^2$
$=3\big(\text{x+2}\big)^2$
View full question & answer→Question 244 Marks
Differentiate the following functions with respect to x:$\frac{\text{a}+\sin\text{x}}{1+\text{a}\sin\text{x}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}+\sin\text{x}}{1+\text{a}\sin\text{x}}\Big)$
Using quotient rule, we get
$\frac{(1+\text{a}\sin\text{x})\frac{\text{d}}{\text{dx}}(\text{a}+\sin\text{x})-(\text{a}+\sin\text{x})\frac{\text{d}}{\text{dx}}(1+\text{a}\sin\text{x})}{(1+\text{a}\sin\text{x})^2}$
$=\frac{(1+\text{a}\sin\text{x})\cos\text{x}-(\text{a}+\sin\text{x})\text{a}\cos\text{x}}{(1+\text{a}\sin\text{x})^2}$
$=\frac{\cos\text{x}+\text{a}\sin\text{x}\cos\text{x}-\text{a}^2\cos\text{x}-\text{a}\sin\text{x}\cos\text{x}}{(1+\text{a}\sin\text{x})^2}$
$=\frac{(1-\text{a}^2)\cos\text{x}}{(1+\text{a}\sin\text{x})^2}$
View full question & answer→Question 254 Marks
Differentiate the following function with respect to x: $(\text{x}\sin\text{x}+\cos\text{x})(\text{x}\cos\text{x}-\sin\text{x})$
Answer$\text{u}=(\text{x}\sin\text{x}+\cos\text{x});\text{v}=(\text{x}\cos\text{x}-\sin\text{x})$$\text{u}'=\text{x}\cos\text{x}+\sin\text{x}-\sin\text{x}=\text{x}\cos\text{x}$
$\text{v}'=-\text{x}\sin\text{x}+\cos\text{x}-\cos\text{x}=-\text{x}\sin\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}=[(\text{x}\sin\text{x}+\cos\text{x})(\text{x}\cos\text{x}-\sin\text{x})]$
$=(\text{x}\sin\text{x}+\cos\text{x})(-\text{x}\sin\text{x})+(\text{x}\cos\text{x}-\sin\text{x})(\text{x}\cos\text{x})$
$=-\text{x}^2\sin^2\text{x}-\text{x}\cos\text{x}\sin\text{x}+\text{x}^2\cos^2\text{x}-\text{x}\cos\text{x}\sin\text{x}$
$=\text{x}^2(\cos^2\text{x}-\sin^2\text{x})-\text{x}(2\sin\text{x}\cos\text{x})$
$=\text{x}^2\cos(2\text{x})-\text{x}(\sin(\text{2x}))$
$=\text{x}[\text{x}\cos(\text{2x})-\sin(\text{2x})]$
View full question & answer→Question 264 Marks
Differentiate the following from first principle:$\frac{\text{x}^2+1}{\text{x}}$
AnswerWe have, $\text{f}\text{(x)}=\frac{\text{x}^2+1}{\text{x}}$ $\because\text{f}'\text{(x)}={\lim\limits_{\text{h}\rightarrow0}}\frac{\text{f(x+}h)-\text{f(x)}}{\text{h}}$ $={\lim\limits_{\text{h}\rightarrow0}}\frac{\frac{\text{(x+h)}^2+1}{\text{(x+h)}}-\frac{\text{x}^2+1}{\text{x}}}{\text{x}}$ $={\lim\limits_{\text{h}\rightarrow0}}\frac{\text{x}\big[\text{x}^2+\text{h}^2+2\text{xh}+1\big]-\big(\text{x}^2+1\big)\big(\text{x+h}\big)}{\text{hx}\big(\text{x+h}\big)}$ $={\lim\limits_{\text{h}\rightarrow0}}\frac{\text{x}^3+\text{xh}^2+2\text{x}^2\text{h}+\text{x}-\text{x}^3-\text{x}-\text{x}^2\text{h}-\text{h}}{\text{h}.\text{x}\text{(x+h)}}$ $={\lim\limits_{\text{h}\rightarrow0}}\frac{\text{xh}+2\text{x}^2-\text{x}^2-1}{\text{x(x+h)}}$ $=\frac{\text{x}^2-1}{\text{x}^2}$ $=1-\frac{1}{\text{x}^2}$
View full question & answer→Question 274 Marks
Differentiate the following functions with respect to x: $\frac{\text{ax}+\text{b}}{\text{px}^2+\text{qx}+\text{r}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{ax}+\text{b}}{\text{px}^2+\text{qx}+\text{r}}\Big)$
$=\frac{(\text{px}^2+\text{qx}+\text{r})\frac{\text{d}}{\text{dx}}(\text{ax}+\text{b})-(\text{ax}+\text{b})\frac{\text{d}}{\text{dx}}(\text{px}^2+\text{qx}+\text{r})}{(\text{px}^2+\text{qx}+\text{r})^2}$
$=\frac{(\text{px}^2+\text{qx}+\text{r})(\text{a})-(\text{ax}+\text{b})(\text{2xp}+\text{q})}{(\text{px}^2+\text{qx}+\text{r})^2}$
$=\frac{(\text{apx}^2+\text{aqx}+\text{ar})-(\text{2apx}^2+\text{aqx}+\text{2bpq}+\text{bq})}{(\text{px}^2+\text{qx}+\text{r})^2}$
$=\frac{-(\text{apx}^2+\text{2bpx}+\text{bq}-\text{ar})}{(\text{px}^2+\text{qx}+\text{r})^2}$
View full question & answer→Question 284 Marks
Differentiate the following from first principle: $\frac{1}{\sqrt{3-\text{x}}}$
AnswerWe have,$\text{f(x)}=\frac{1}{\sqrt{{3}-\text{x}}}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\sqrt{3-(\text{x+h)}}}-\frac{1}{\sqrt{3-\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{3-\text{x}}-\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{3-\text{x}}-\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}\sqrt{3-\text{(x+h)}\text{h}}}\times\frac{\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}}$
$\Big[ $Rationalising the numerator by $\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(3-\text{x})-\big(3-\text{(x+h)}\big)}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}\bigg(\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}\bigg)}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}\big(\sqrt{3-\text{x}}+{\sqrt{3-\text{(x+h)}}}\big)}}$
$=\frac{1}{(3-\text{x})\times2\sqrt{3-\text{x}}}$
$=\frac{1}{2(3-\text{x})^\frac{3}{2}}$
View full question & answer→Question 294 Marks
Differentiate the following functions by the product rule and the other method and verify that the answer from both the methods is the same.$(3\sec\text{x}-4\text{cosec}\text{x})(-2\sin\text{x}+5\cos\text{x})$
AnswerLet $\text{u}=(3\sec\text{x}-4\text{cosec}\text{x});\text{v}=(-2\sin\text{x}+5\cos\text{x})$Then, $\text{u}'=3\sec\text{x}\tan\text{x}+4\text{cosec}\text{x}\cot\text{x};\text{v}'=-2\cos\text{x}-5\sin\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(3\sec\text{x}-4\text{cosec}\text{x})(-2\sin\text{x}+5\cos\text{x})]$
$=(3\sec\text{x}-4\text{cosec}\text{x})(-2\cos\text{x}-5\cos\text{x})+(-2\sin\text{x}+5\cos\text{x})(3\sec\text{x}\tan\text{x}+4\text{cosec}\text{x}\cot\text{x})$
$=-6+15\tan\text{x}+8\cot\text{x}+20-6\tan^2\text{x}-8\cot\text{x}-15\tan\text{x}+20\cot^2\text{x}$
$=-6+20-6(\sec^2\text{x}-1)+20(\text{cosec}^2\text{x}-1)$
$=-6\sec^2\text{x}+20\text{cosec}^2\text{x}$
Alternate method
$\frac{\text{d}}{\text{dx}}[(3\sec\text{x}-4\text{cosec}\text{x})(-2\sin\text{x}+5\cos\text{x})]$
$=\frac{\text{d}}{\text{dx}}(-6\sec\text{x}\sin\text{x}+15\sec\text{x}\cos\text{x}+8\text{cosecx}\sin\text{x}-20\text{cosecx}\cos\text{x})$
$=\frac{\text{d}}{\text{dx}}\Big(-6\frac{\sin\text{x}}{\cos\text{x}}+15\frac{\cos\text{x}}{\cos\text{x}}+8\frac{\sin\text{x}}{\sin\text{x}}-20\frac{\cos\text{x}}{\sin\text{x}}\Big)$
$=\frac{\text{d}}{\text{dx}}(-6\tan\text{x}+15+8-20\cot\text{x})$
$=\frac{\text{d}}{\text{dx}}(-6\tan\text{x}-20\cot\text{x}+23)$
$-6\sec^2\text{x}+20\text{cosec}^2\text{x}$
View full question & answer→Question 304 Marks
Differentiate the following from first principle:$\text{e}^{\text{ax}+\text{b}}$
AnswerWe have,$\text{f(x)}=\text{e}^{\text{ax}+\text{b}}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{a}\text{(x+h)}+\text{b}}-\text{e}^{\text{ax+b}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{\text{ax}}\times\text{e}^{\text{ah}}\times\text{e}^\text{b}-\text{e}^{\text{ax}}\times\text{e}^\text{b}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{b}\times\text{e}^{\text{ax}}\big(\text{e}^{\text{ax}}-1\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\text{ax+b}}\times\frac{\text{a}\big(\text{e}^{\text{ah}}-1\big)}{\text{a.h}}$
Multiplying Numerator and denominator by a $\bigg[=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{e}^{\text{ah}}-1\big)}{\text{ah}}=1\bigg]$
$\text{ae}^{\text{ax+b}}$
View full question & answer→Question 314 Marks
Differentiate the following from first principle:$\frac{1}{\text{x}^3}$
AnswerWe have,$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\text{(x+h)}^3}-{\frac{1}{\text{x}^3}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^3-\text{(x+h)}^3}{\text{x}^3\text{h(x+h)}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^3-\text{(x}^3+3\text{x}^2\text{h}+3{\text{xh}^2+\text{h}^3)}}{\text{x}^3\text{h}\text{(x+h)}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^3-\text{x}^3-3\text{x}^2-3\text{xh}-\text{h}^2}{\text{x}^3\text{(x+h)}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-3\text{x}^2-3\text{xh}-\text{h}^2}{\text{x}^3\text{(x+h)}^3}$
$=\frac{-3\text{x}^2}{\text{x}^6}$
$=\frac{-3}{\text{x}^4}$
View full question & answer→Question 324 Marks
Differentiate the following from first principle:$\text{e}^{-\text{x}}$
AnswerWe have, $\text{f(x)}=\text{e}^{-\text{x}}$ $\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\big(\text{x+h}\big)-\text{f}\big(\text{x}\big)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{-(\text{x+h})}-\text{e}^{-\text{x}}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{-\text{x}}\big(\text{e}^{-\text{h}}-1\big)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{e}^{-\text{x}}\big(\text{e}^{-\text{h}}-1\big)}{-\text{h}}$ $= -\text{e}^{-\text{x}}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\text{e}^{\theta}-1}{\theta}=1\Big]$
View full question & answer→Question 334 Marks
Differentiate the following from the first principle$\sin(\text{x}+1)$
AnswerLet $\text{f}(\text{x})=\sin(\text{x}+1).$ Then, $\text{f}(\text{x}+1)=\sin\big((\text{x}+\text{h})+1\big)$$\therefore\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\big((\text{x}+\text{h})+1\big)-\sin(\text{x}+1)}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\Bigg[\frac{\big((\text{x}+\text{h})+1\big)-(\text{x}+1)}{2}\Bigg]\cos\Bigg[\frac{\big((\text{x}+\text{h})+1\big)+(\text{x}+1)}{2}\Bigg]}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\Big[\frac{\text{h}}{2}\Big]\cos\Big[\frac{2\text{x}+2+\text{h}}{2}\Big]}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big[\frac{\text{h}}{2}\Big]}{\frac{\text{h}}{2}}\times\lim_\limits{\text{h}\rightarrow0}\cos\Big[\frac{2\text{x}+2+\text{h}}{2}\Big]$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=1\times\cos\Big[\frac{\text{2x}+2+0}{2}\Big]$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\cos(\text{x}+1)$
View full question & answer→Question 344 Marks
Differentiate the following from first principle:$\text{x}^3+4\text{x}^2+3\text{x}+2$
AnswerWe have,$\text{f(x)}=\text{x}^3+4\text{x}^2+3\text{x}+2$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x+h})^3+4(\text{x+h})^2+3(\text{x+h})+2-(\text{x}^3+4\text{x}^2+3\text{x}+2)}{\text{h}}$
On solving we get,$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^3+\text{h}^3+3\text{x}^2\text{h}+3\text{h}^2\text{x}+4\text{x}^2+4\text{h}^2+8\text{hx}+3\text{x}+3\text{h}+2-\text{x}^3-4\text{x}^2-3\text{x}-2}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{3\text{x}^2\text{h}+3\text{xh}^2+\text{h}^3+4\text{h}^2+8\text{hx}+3\text{h}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}3\text{x}^2+3\text{xh}+\text{h}^2+4\text{h}+8\text{x}+3$
$=3\text{x}^2+8\text{x}+3$
View full question & answer→Question 354 Marks
Differentiate the following functions with respect to x:$\frac{\text{x}+\cos\text{x}}{\tan\text{x}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}+\cos\text{x}}{\tan\text{x}}\Big)$
Using quotient rule, we get
$\frac{(\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+\cos\text{x})-(\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\tan\text{x})}{(\tan^2\text{x})}$
$=\frac{\tan\text{x}\Big\{1+(-\sin\text{x})\Big\}-(\text{x}+\cos\text{x})\sec^2\text{x}}{(\tan^2\text{x})}$
$=\frac{(1-\sin\text{x})\tan\text{x}-(\text{x}+\cos\text{x})\sec^2\text{x}}{(\tan^2\text{x})}$
View full question & answer→Question 364 Marks
Differentiate the following function with respect to x:$\frac{(\text{ax}+\text{b})}{(\text{cx}+\text{d})}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}\Big)=\frac{(\text{cx}+\text{d})\frac{\text{d}}{\text{dx}}(\text{ax}+\text{b})-(\text{ax}+\text{b})\frac{\text{d}}{\text{dx}}(\text{cx}+\text{d})}{(\text{cx}+\text{d})^2}$ (Using divident rule)$=\frac{(\text{cx}+\text{d})\text{a}-(\text{ax}+\text{b})\text{c}}{(\text{cx}+\text{d})^2}$
$=\frac{(\text{axc}+\text{ad}-\text{axc}-\text{bc})}{(\text{cx}+\text{d})^2}$
$=\frac{\text{ad}-\text{bc}}{(\text{cx}+\text{d})^2}$
View full question & answer→Question 374 Marks
Differentiate the following from the first principle$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{x}}$
AnswerWe have, $\text{f}(\text{x})=\frac{\sin\text{x}}{\text{x}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})}{\text{x}+\text{h}}-\frac{\sin\text{x}}{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\sin(\text{x}+\text{h})-(\text{x}+\text{h})\sin\text{x}}{\text{xh}(\text{x}+\text{h})}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}(\sin\text{x}.\cos\text{h}+\cos\text{h}.\sin\text{h})-\text{x}\sin\text{x}-\text{h}.\sin\text{x}}{\text{xh}(\text{x}+\text{h})}$ $[\because\sin(\text{A}+\text{B}=\sin\text{A}.\cos{B}+\cos\text{A}.\sin\text{B})]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}.\sin\text{x}(\cos\text{h}-1)}{\text{xh}(\text{x}+\text{h})}+\frac{\text{x}\cos\text{x}.\sin\text{x}}{\text{xh}(\text{x}+\text{h})}-\frac{\text{h}\sin\text{x}}{\text{xh}(\text{x}+\text{h})}$ $\Big[\because1-\cos\text{h}=2\sin^2\frac{\text{h}}{2}\Big]$
$=\frac{-\text{x}\sin\text{x}}{\text{x}(\text{x}+\text{h})}\times\frac{2\sin^2\frac{\text{h}}{2}}{\frac{\text{h}^2}{4}}\times\frac{\text{h}}{4}+\frac{\text{x}\cos\text{x}}{\text{x}^2}-\frac{\sin\text{x}}{\text{x}^2}$
$\because\text{h}\rightarrow0\Rightarrow\frac{\text{h}}{2}\rightarrow0$ and $\lim_\limits{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1$
$=0+\frac{\text{x}\cos\text{x}-\sin\text{x}}{\text{x}^2}$
$=\frac{\text{x}\cos\text{x}-\sin\text{x}}{\text{x}^2}$
View full question & answer→Question 384 Marks
Differentiate the following functions with respect to x:$\frac{1+3^\text{x}}{1-3^\text{x}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{1+3^\text{x}}{1-3^\text{x}}\Big)$
Using quotient rule, we get
$\frac{(1-3^\text{x})\frac{\text{d}}{\text{dx}}(1+3^\text{x})-(1+3^\text{x})\frac{\text{d}}{\text{dx}}(1-3^\text{x})}{(1-3^\text{x})^2}$
$=\frac{(1-3^\text{x})3^\text{x}\log3+(1+3^\text{x})3^\text{x}\log3}{(1-3^\text{x})^2}$
$=\frac{3^\text{x}\log3-3^\text{x}\times3^\text{x}\log3+3^\text{x}\log3+3^\text{x}\times3^\text{x}\log3}{(1-3^\text{x})^2}$
$=\frac{2\times3^\text{x}\log3}{(1-3^\text{x})^2}$
View full question & answer→Question 394 Marks
Differentiate the following function with respect to x:$\text{x}^4(5\sin\text{x}-3\cos\text{x})$
AnswerLet $\text{u}=\text{x}^4;\text{v}=5\sin\text{x}-3\cos\text{x}$Then, $\text{u}'=\text{4x}^3;\text{v}'=5\cos\text{x}-3(-\sin\text{x})=5\cos\text{x}+3\sin\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{u}'\text{v}+\text{uv}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^4(5\sin\text{x}-3\cos\text{x}))=\text{x}^4(5\cos\text{x}+3\sin\text{x})+\text{4x}^3(5\sin\text{x}-3\cos\text{x})$
$=\text{x}^3(\text{5x}\cos\text{x}+\text{3x}\sin\text{x}+20\sin\text{x}-12\cos\text{x})$
$=\text{x}^3((\text{3x}+20)\sin\text{x}+(\text{5x}-12)\cos\text{x})$
$=\text{3x}^4\sin\text{x}+20\text{x}^3\sin\text{x}+\text{5x}\cos\text{x}-12\cos\text{x}$
View full question & answer→Question 404 Marks
Differentiate the following function with respect to x:$\log_{\text{x}^2}\text{x}$
Answer$\log_{\text{x}^2}\text{x}=\frac{\log\text{x}}{\log\text{x}^2}$ (By change of base property)$=\frac{\log\text{x}}{2\log\text{x}}[\log\text{x}^2=2\log\text{x}]$
$=\frac{1}{2}$
Now $\frac{\text{d}}{\text{dx}}(\log_{\text{x}^2}\text{x})=\frac{\text{d}}{\text{dx}}\Big(\frac{1}{2}\Big)$
$=0(\because\frac{1}{2}$ is a constant)
View full question & answer→Question 414 Marks
Differentiate the following functions with respect to x:$\frac{\text{x}\sin\text{x}}{1+\cos\text{x}}$
AnswerWe have, $\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}\sin\text{x}}{1+\cos\text{x}}\Big)$
Using Quotient rule, we get
$=\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{x})-(\text{x}\sin\text{x})\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}$
$=\frac{(1+\cos\text{x})\Big(\text{x}\frac{\text{d}}{\text{dx}}\sin\text{x}+\sin\text{x}\frac{\text{d}}{\text{dx}}\Big)-\text{x}\sin\text{x}(-\sin\text{x})}{(1+\cos\text{x})^2}$ [Used product rule]
$=\frac{(1+\cos\text{x})(\text{x}\cos\text{x}+\sin\text{x})+\text{x}\sin^2\text{x}}{(1+\cos\text{x})^2}$
$=\frac{\text{x}\cos\text{x}+\text{x}\cos^2\text{x}+\sin\text{x}+\sin\text{x}\cos\text{x}+\text{x}\sin^2\text{x}}{(1+\cos\text{x})^2}$
$=\frac{(\text{x}\cos\text{x}+\sin\text{x}+\sin\text{x}\cos\text{x})+\text{x}(\sin^2\text{x}+\cos^2\text{x})}{(1+\cos\text{x})^2}$
$=\frac{\sin\text{x}+\sin\text{x}\cos\text{x}+\text{x}(\cos\text{x}+\sin^2\text{x}+\cos^2\text{x})}{(1+\cos\text{x})^2}$
$=\frac{\sin\text{x}(1+\cos\text{x})+\text{x}(\cos\text{x}+1)}{(1+\cos\text{x})^2}$
$=\frac{(\text{x}+\sin\text{x})(\cos\text{x}+1)}{(1+\cos\text{x})^2}$
View full question & answer→Question 424 Marks
Differentiate the following from first principle:$\sqrt{2\text{x}^2-1}$
AnswerWe have, $\text{f(x)}=\sqrt{2\text{x}^2+1}$ $\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2(\text{x+h})^2+1}-\sqrt{2\text{x}^2+1}}{\text{h}}$ Multiplying Numerator and denominator by ${\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg\{2(\text{x+h})^2+1-(2\text{x}^2+1)\bigg\}}{\text{h}\bigg(\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}\bigg)}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{x}^2+2\text{h}^2+4\text{xh}+1-2\text{x}^2-1}{\text{h}\bigg(\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}\bigg)}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{4\text{xh}+2\text{h}^2}{\text{h}\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}}$ $=\frac{4\text{x}}{2\sqrt{2\text{x}^2+1}}$ $=\frac{2\text{x}}{\sqrt{2\text{x}^2+1}}$ $$
View full question & answer→Question 434 Marks
Differentiate the following from the first principle$\sqrt{\sin(\text{3x}+1)}$
AnswerWe have, $\text{f}(\text{x})=\sqrt{\sin(\text{3x}+1)}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin3(\text{x}+\text{h})+1}-\sqrt{\sin(\text{3x}+1)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin(\text{3x}+\text{3h})+1}-\sqrt{\sin(\text{3x}+1)}}{\text{h}}\times\frac{{\sqrt{\sin(\text{3x}+\text{3h})+1}+\sqrt{\sin(\text{3x}+1)}}}{{\sqrt{\sin(\text{3x}+\text{3h})+1}+\sqrt{\sin(\text{3x}+1)}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{3x}+\text{3h}+1)-\sin(\text{3x}+1)}{\text{h}\big(\sqrt{\sin(\text{3x}+\text{3h})+1}+\sqrt{\sin(\text{3x}+1)}\big)}$
$=\lim_\limits{\text{h}\rightarrow0}2\cos\Big(\text{3x}+1+\frac{\text{3h}}{2}\Big)\times\frac{\sin\frac{\text{3h}}{2}}{\frac{\text{3h}}{2}}\times\frac{3}{2}\times\frac{1}{\sqrt{\sin(\text{3x}+\text{3h})+1}+\sqrt{\sin(\text{3x}+1)}}$
$=\frac{3\cos(\text{3x}+1)}{2\sqrt{\sin(\text{3x}+1)}}$ $\Bigg[\lim_\limits{\text{h}\rightarrow0}\frac{\sin\frac{\text{3h}}{2}}{\frac{\text{3h}}{2}}=1\Bigg]$
View full question & answer→Question 444 Marks
Differentiate the following from first principle$\tan(\text{2x}+1)$
Answer$\text{f}(\text{x})=\tan(\text{2x}+1)$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan\big\{2(\text{x}+\text{h})+1\big\}-\tan(\text{2x}+1)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}+1-\text{2x}-1)}{\text{h}.\cos\big\{2(\text{x}+\text{h})+1\big\}\cos(\text{2x}+1)}$
$\Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin(\text{A}-\text{B})}{\cos\text{A}.\cos\text{B}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin2\text{h}}{2\text{h}.\cos(\text{2x}+\text{2h}+1)\cos(\text{2x}+1)}$
Multiplying both, numerator and denominator by 2.
$\therefore\lim_\limits{\text{h}\rightarrow0}2\Big(\frac{\sin2\text{h}}{2}\Big)\times\frac{1}{\cos(\text{2x}+\text{2h}+1)\cos(\text{2x}+1)}$
$=\frac{2}{\cos^2(\text{2x}+1)}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2}=1\Big]$
$=2\sec^2(\text{2x}+1)\ \Big[\because\sec^2\text{x}=\frac{1}{\cos^2\text{x}}\Big]$
$=2\sec^ 2(\text{2x}+1)$
View full question & answer→Question 454 Marks
Differentiate the following function with respect to $(\text{x})$:$\Big\{\log\Big(\frac{1}{\sqrt{\text{x}}}\Big)+5\text{x}^\text{a}-\text{3a}^\text{x}+\sqrt[3]{\text{x}^2}+6\sqrt[4]{\text{x}^{-3}}\Big\}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big\{\log\Big(\frac{1}{\sqrt{\text{x}}}\Big)+5\text{x}^\text{a}-\text{3a}^\text{x}+\sqrt[3]{\text{x}^2}+6\sqrt[4]{\text{x}^{-3}}\Big\}$
$=\frac{\text{d}}{\text{dx}} \log\Big(\frac{1}{\sqrt{\text{x}}}\Big)+5\frac{\text{d}}{\text{dx}}(\text{x}^\text{a})-3(\text{a}^\text{x})+\frac{\text{d}}{\text{dx}}(\sqrt[3]{\text{x}^2})+6\frac{\text{d}}{\text{dx}}(\sqrt[4]{\text{x}^{-3}})$
$=\frac{-1}{2}\frac{1}{\text{x}}+5\text{ax}^{\text{a}-1}-3\text{a}^\text{x}\log\text{a}+\frac{2\text{x}^{\frac{-1}{3}}}{3}+\text{6x}^{\frac{-7}{4}}\Big(\frac{-3}{4}\Big)$
$=\frac{-1}{2\text{x}}+5\text{ax}^{\text{a}-1}-3\text{a}^\text{x}\log\text{a}+\frac{2\text{x}^{\frac{-1}{3}}}{3}-\frac{9}{2}\text{x}^{\frac{-7}{4}}$
View full question & answer→Question 464 Marks
Differentiate the following from the first principle$\text{f}(\text{x})=\frac{\cos\text{x}}{\text{x}}$
AnswerWe have, $\text{f}(\text{x})=\frac{\cos\text{x}}{\text{x}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\cos(\text{x}+\text{h})}{\text{x}+\text{h}}-\frac{\cos\text{x}}{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\cos(\text{x}+\text{h})-(\text{x}+\text{h})\cos\text{x}}{\text{xh}(\text{x}+\text{h})}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}(\cos\text{x}.\cos\text{h}+\sin\text{h}.\sin\text{h})-\text{x}\cos\text{x}-\text{h}.\cos\text{x}}{\text{xh}(\text{x}+\text{h})}$ $[\because\cos(\text{A}+\text{B}=\cos\text{A}.\cos{B}+\sin\text{A}.\sin\text{B})]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}.\cos\text{x}(\cos\text{h}-1)}{\text{xh}(\text{x}+\text{h})}+\frac{\text{x}\cos\text{x}.\sin\text{h}}{\text{xh}(\text{x}+\text{h})}-\frac{\text{h}\cos\text{x}}{\text{xh}(\text{x}+\text{h})}$
$=\frac{-\text{x}\cos\text{x}}{\text{x}(\text{x}+\text{h})}\times\frac{2\sin^2\frac{\text{h}}{2}}{\frac{\text{h}^2}{4}}\times\frac{\text{h}^2}{4}-\frac{\text{x}\sin\text{x}}{\text{x}(\text{x}+\text{h})}-\frac{\cos\text{x}}{\text{x}(\text{x}+\text{h})}$
$=0-\frac{\text{x}\sin\text{x}}{\text{x}^2}-\frac{\cos\text{x}}{\text{x}^2}$
$=\frac{-\text{x}\sin\text{x}-\cos\text{x}}{\text{x}^2}$
View full question & answer→Question 474 Marks
Differentiate the following from first principle:$\text{x}\text{e}^\text{x}$
AnswerWe have,$\text{f(x)}=\text{xe}^\text{x}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x+h})\text{e}^{(\text{x+h})}-\text{xe}^\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{xe}^\text{x}.\text{e}^\text{h}+\text{he}^\text{x}.\text{e}^\text{h}-\text{xe}^\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{xe}^\text{x}\Big(\frac{\text{e}^\text{h}-1}{\text{h}}\Big)+\frac{\text{he}^{\text{x}+\text{h}}}{\text{h}}$
$=\text{xe}^\text{x}+\text{e}^\text{x}$
$=\text{e}^\text{x}\big(\text{x}+1\big)$
View full question & answer→Question 484 Marks
Differentiate the following functions with respect to x:$\frac{\text{x}}{1+\tan\text{x}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{1+\tan\text{x}}\Big)$$=\frac{(1+\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(1+\tan\text{x})}{(1+\tan\text{x})^2}$
$=\frac{(1+\tan\text{x})-\text{x}(\sec^2\text{x})}{(1+\tan\text{x})^2}$
$=\frac{1+\tan\text{x}-\text{x}\sec^2\text{x}}{(1+\tan\text{x})^2}$
View full question & answer→Question 494 Marks
Differentiate the following functions with respect to x:$\frac{\sin\text{x}-\text{x}\cos\text{x}}{\text{x}\sin\text{x}+\cos\text{x}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}-\text{x}\cos\text{x}}{\text{x}\sin\text{x}+\cos\text{x}}\Big)$
Using quotient rule, we get
$\frac{(\text{x}\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\text{x}\cos\text{x})-(\sin\text{x}-\text{x}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{x}+\cos\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})\Big\{\cos\text{x}-\Big(\frac{\text{dx}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{dx}}{\text{dx}}\Big)\Big\}-(\sin\text{x}-\text{x}\cos\text{x})\Big(\frac{\text{dx}}{\text{dx}}\sin\text{x}+\sin\text{x}\frac{\text{dx}}{\text{dx}}\Big)+\frac{\text{d}}{\text{dx}}\cos\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})(\cos\text{x}+\text{x}\sin\text{x}-\cos\text{x})-(\sin\text{x}-\text{x}\cos\text{x})(\text{x}\cos\text{x}+\sin\text{x}-\sin\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})\text{x}\sin\text{x}-(\sin\text{x}-\text{x}\cos\text{x})\text{x}\cos\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{\text{x}^2\sin^2\text{x}+\text{x}\sin\text{x}\cos\text{x}-\text{x}\sin\text{x}\cos\text{x}+\text{x}^2\cos^2\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{\text{x}^2(\sin^2\text{x}+\cos^2\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}\ (\because\sin^2\text{x}+\cos^2\text{x}=1)$
$=\frac{\text{x}^2}{\text{x}\sin\text{x}+\cos\text{x}}$
View full question & answer→Question 504 Marks
Differentiate the following functions with respect to x:$\frac{\text{4x}+5\sin\text{x}}{\text{3x}+7\cos\text{x}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{4x}+5\sin\text{x}}{\text{3x}+7\cos\text{x}}\Big)$
Using quotient rule, we get
$\frac{(\text{3x}+7\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{4x}+5\sin\text{x})-(4\text{x}+5\sin\text{x})\frac{\text{d}}{\text{dx}}(\text{3x}+7\cos\text{x})}{(\text{3x}+7\cos\text{x})^2}$
$=\frac{(\text{3x}+7\cos\text{x})(4+5\cos\text{x})-(4\text{x}+5\sin\text{x})(3+7(-\sin\text{x}))}{(\text{3x}+7\cos\text{x})^2}$
$=\frac{12\text{x}+28\cos\text{x}+\text{15x}\cos\text{x}+13\cos^2\text{x}-\text{12x}-15\sin\text{x}+\text{28x}\sin\text{x}+25\sin^2\text{x}}{(\text{3x}+7\cos\text{x})^2}$
$=\frac{\text{15x}\cos\text{x}+\text{28x}\sin\text{x}+28\cos\text{x}-15\sin\text{x}+35(\sin^2\text{x}+\cos^2\text{x})}{(\text{3x}+7\cos\text{x})^2}$
$\therefore\frac{\text{15x}\cos\text{x}+\text{28x}\sin\text{x}+28\cos\text{x}-15\sin\text{x}+35}{(\text{3x}+7\cos\text{x})^2}$
View full question & answer→Question 514 Marks
Differentiate the following functions by the product rule and the other method and verify that the answer from both the methods is the same.$(\text{x}+2)(\text{x}+3)$
AnswerLet $\text{u}=(\text{x}+2);\text{v}=(\text{x}+3)$Then, $\text{u}'=1;\text{v}'=1$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(\text{x}+2)(\text{x}+3)]=(\text{x}+2)1+(\text{x}+3)1$
$=\text{x}+2+\text{x}+3$
$=\text{2x}+5$
Alternate method
$\frac{\text{d}}{\text{dx}}[(\text{x}+2)(\text{x}+3)]$
$=\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{5x}+6)$
$=\text{2x}+5$
View full question & answer→Question 524 Marks
Differentiate the following from the first principle$\text{x}\cos\text{x}$
Answerwe have, $\text{f}(\text{x})=\text{x}\cos\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})\cos(\text{x}+\text{h})-\text{x}\cos\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\cos(\text{x}+\text{h})\text{h}\cos(\text{x}+\text{h})-\text{x}\cos\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}\Big\{\cos(\text{x}+\text{h})-\cos\text{x}\Big\}}{\text{h}}+\cos(\text{x}+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{x}.2\sin\Big(\text{x}-\text{x}+\frac{\text{h}}{2}\Big)\sin\Big(\text{x}+\frac{\text{h}}{2}\Big)+\cos(\text{x}+\text{h})$ $\Big[\because\cos\text{}A-\cos\text{B}=2\sin\frac{\text{B}-\text{A}}{2}\sin\frac{\text{B}+\text{A}}{2}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}2\text{x}.\sin\Big[\frac{-\text{h}}{2}\Big]\sin\Big(\text{x}+\frac{\text{h}}{2}\Big)+\cos(\text{x}+\text{h})$
$=-\text{x}\sin\text{x}+\cos\text{x}$
View full question & answer→Question 534 Marks
Differentiate the following functions with respect to x:$\frac{\text{x}^5-\cos\text{x}}{\sin\text{x}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^5-\cos\text{x}}{\sin\text{x}}\Big)$
Using quotient rule, we get
$\frac{(\sin\text{x})\frac{\text{d}}{\text{dx}}(\text{x}^5-\cos\text{x})-(\text{x}^5-\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x})}{(\sin^2\text{x})}$
$=\frac{\sin\text{x}(\text{5x}^4\sin\text{x})-(\text{x}^5-\cos\text{x})\cos\text{x}}{(\sin^2\text{x})}$
$=\frac{\text{5x}^4\sin\text{x}+\sin^2\text{x}-\text{x}^5\cos\text{x}+\cos^2\text{x}}{(\sin^2\text{x})}\ (\because\sin^2\text{x}+\cos^2\text{x}=1)$
$=\frac{-\text{x}^5\cos\text{x}+\text{5x}^4\sin\text{x}+1}{(\sin^2\text{x})}$
View full question & answer→Question 544 Marks
Differentiate the following functions with respect to x:$\frac{\sin\text{x}-\text{x}\cos\text{x}}{\text{x}\sin\text{x}+\cos\text{x}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}-\text{x}\cos\text{x}}{\text{x}\sin\text{x}+\cos\text{x}}\Big)$
Using quotient rule, we get
$\frac{(\text{x}\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\text{x}\cos\text{x})-(\sin\text{x}-\text{x}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{x}+\cos\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})\Big\{\cos\text{x}-\Big(\frac{\text{dx}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{dx}}{\text{dx}}\Big)\Big\}-(\sin\text{x}-\text{x}\cos\text{x})\Big(\frac{\text{dx}}{\text{dx}}\sin\text{x}+\sin\text{x}\frac{\text{dx}}{\text{dx}}\Big)+\frac{\text{d}}{\text{dx}}\cos\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})(\cos\text{x}+\text{x}\sin\text{x}-\cos\text{x})-(\sin\text{x}-\text{x}\cos\text{x})(\text{x}\cos\text{x}+\sin\text{x}-\sin\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{(\text{x}\sin\text{x}+\cos\text{x})\text{x}\sin\text{x}-(\sin\text{x}-\text{x}\cos\text{x})\text{x}\cos\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{\text{x}^2\sin^2\text{x}+\text{x}\sin\text{x}\cos\text{x}-\text{x}\sin\text{x}\cos\text{x}+\text{x}^2\cos^2\text{x}}{(\text{x}\sin\text{x}+\cos\text{x})^2}$
$=\frac{\text{x}^2(\sin^2\text{x}+\cos^2\text{x})}{(\text{x}\sin\text{x}+\cos\text{x})^2}\ (\because\sin^2\text{x}+\cos^2\text{x}=1)$
$=\frac{\text{x}^2}{\text{x}\sin\text{x}+\cos\text{x}}$
View full question & answer→Question 554 Marks
Differentiate the following from the first principle$\text{e}^{\sqrt{\text{ax}+\text{b}}}$
AnswerWe have, $\text{f}(\text{x})=\text{e}^{\sqrt{\text{ax}+\text{b}}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}}-\text{e}^{\sqrt{\text{ax}+\text{b}}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\text{a}(\text{x}+\text{h}+\text{b})}}\big(\text{e}^{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}-1\big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\text{a}(\text{x}+\text{h}+\text{b})}}\big(\text{e}^{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}-1\big)}{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}\times\frac{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}{\text{h}}$
Multiplying numerator and denominator by $\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}$
$\because\text{h}\rightarrow0\Rightarrow\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}$
and $\lim_\limits{\theta\rightarrow0}\frac{\text{e}^\theta-1}{\theta}=1$
$\therefore\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{ax}+\text{b}}}\times1\times\frac{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}}\times\frac{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}+\sqrt{\text{ax}+\text{b}}}{\text{h}}$
Again Multiplying numerator and denominator by $\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}-\sqrt{\text{ax}+\text{b}}$
$\therefore\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{ax}+\text{b}}}\times\frac{{\text{a}(\text{x}+\text{h})+\text{b}}-{\text{ax}+\text{b}}}{\text{h}}\times\frac{1}{\sqrt{\text{a}(\text{x}+\text{h})+\text{b}}+\sqrt{\text{ax}+\text{b}}}$
$=\frac{\text{e}^{\sqrt{\text{ax}+\text{b}}}\times\text{a}}{2\sqrt{\text{ax}+\text{b}}}$
$=\frac{\text{ae}^{\sqrt{\text{ax}+\text{b}}}}{2\sqrt{\text{ax}+\text{b}}}$
View full question & answer→Question 564 Marks
Differentiate the following from first principle$\text{a}^{\sqrt{\text{x}}}$
Answer$\text{f}(\text{x})=\text{a}^\sqrt{\text{x}}=\text{e}^{\sqrt{\text{x}}\log\text{a}}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\text{x}+\text{h}}\log\text{a}}-\text{e}^{\sqrt{\text{x}}\log\text{a}-1}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{x}}\log\text{a}}\frac{\text{e}^{\sqrt{\text{x}+\text{h}}\log\text{a}-\sqrt{\text{x}}\log\text{a}}-1}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{x}}\log\text{a}}\frac{\text{e}^{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}-1}{\text{h}}$
Multiply numerator and denominator by $(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}$
$\text{f}(\text{x})=\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{x}}\log\text{a}}\frac{\text{e}^{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}-1}{\text{h}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}$
$=\text{e}^{\sqrt{\text{x}}\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}-1}{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\log\text{a}\frac{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}{\text{h}}$
$=\text{e}^{\sqrt{\text{x}}\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\log\text{a}\frac{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}{\text{h}}$
Multiply numerator and denominator by $(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})$
$\text{f}(\text{x})=\text{e}^{\sqrt{\text{x}}\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\log\text{a}\frac{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}{\text{h}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})$
$=\text{e}^{\sqrt{\text{x}}\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\log\text{a}\frac{\text{h}}{\text{h}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}$
$=\text{e}^{\sqrt{\text{x}}\log\text{a}}\frac{\log\text{a}}{2\sqrt{\text{x}}}$
$=\frac{\text{a}^{\sqrt{\text{x}}}}{2\sqrt{\text{x}}}\log_\text{e}\text{a}$
View full question & answer→Question 574 Marks
Differentiate the following from first principle:$(-\text{x})^{-1}$
AnswerLet $\text{f}(\text{x})=(-\text{x})^{-1}.$ Then, $\text{f}(\text{x}+\text{h})=\Big(-(\text{x}+\text{h})\Big)^{-1}$$\therefore\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\big(-(\text{x}+\text{h})^{-1}-(-\text{x})^{-1}\big)}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{-1}{\text{x}+\text{h}}+\frac{1}{\text{x}}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{-\text{x}+\text{x}+\text{h}}{\text{x}(\text{x}+\text{h})}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}\text{x}(\text{x}+\text{h})}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{1}{\text{x}(\text{x}+\text{h})}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\frac{1}{\text{x}(\text{x}+0)}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\frac{1}{\text{x}^2}$
View full question & answer→Question 584 Marks
If for $\text{f}(\text{x})=\lambda\text{x}^2+\mu\text{x}+12,\text{f}'(\text{x})=15$ and $\text{f}'(\text{2})=11,$ then find $\lambda$and $\mu.$
AnswerWe have,$\text{f}(\text{x})=\lambda\text{x}^2+\mu\text{x}+12$
$\Rightarrow\text{f}'(\text{x})=\text{2x}\lambda+\mu\dots(\text{i})$
but, $\text{f}'(4)=15$
from (i)
$8\lambda+\mu=15\dots(\text{ii})$
also, $\text{f}'(\text{12})=11$
$4\lambda+\mu=11\dots(\text{iii})$
(ii)-(iii) gives
$4\lambda=4$
$\Rightarrow\lambda=1$
from (ii)
$8.1+\mu=15$
$\Rightarrow\mu=7$
Hence,
$\lambda=1$ and $\mu=7$
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