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Maths Solve the Following Question.(5 Marks)

Derivatives · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 16 questions.

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Solve the Following Question.(5 Marks)

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16 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks
Differentiate the following from first principle$\sin\sqrt{2\text{x}}$
Answer
We 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)$$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})}{2}\Bigg)}{\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})}{2}\Bigg)}\lim_\limits{\text{h}\rightarrow0}\frac{{2(\text{x}+\text{h})}-{2\text{x}}}{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})\text{h}}\lim_\limits{\text{h}\rightarrow0}\cos\Bigg(\frac{\sqrt{2(\text{x}+\text{h})+\text{2x}}}{2}\Bigg)$
$=1\times\frac{2}{2\sqrt{2\text{x}}}\cos(\sqrt{2\text{x}})$
$=\frac{\cos(\sqrt{2\text{x}})}{\sqrt{2\text{x}}}$
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Question 25 Marks
Differentiate the following from first principle$3^{\text{x}^2}$
Answer
$\text{f}(\text{x})=3^{\text{x}^2}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$\frac{\text{d}}{\text{dx}}\big(3^{\text{x}^2}\big)=\lim_\limits{\text{h}\rightarrow0}\frac{3^{(\text{x}+\text{h})^2}-3^{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{3^{\text{x}^2+\text{2xh}+\text{h}^2}-3^{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{3^{\text{x}^2}(3^{\text{x}^2+\text{2xh}+\text{h}^2-\text{x}^2}-1)}{\text{h}}\times\frac{(\text{h}+\text{2x})}{(\text{h}+\text{2h})}$
$=3^{\text{x}^2}\lim_\limits{\text{h}\rightarrow0}\frac{3^{\text{h}(\text{h}+\text{2x})}-1}{\text{h}(\text{h}+\text{2x})}\lim_\limits{\text{h}\rightarrow0}(\text{h}+\text{2x})$
$=3^{\text{x}^2}\log3(2\text{x})$
$=2\text{x}3^{\text{x}^2}\log3$
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Question 35 Marks
Differentiate the following from the first principle$\text{x}^2\sin\text{x}$
Answer
We have, $\text{f}(\text{x})=\text{x}^2\sin\text{x}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})^2\sin(\text{x}+\text{h})-\text{x}^2\sin\text{x}}{\text{h}}$
$\lim_\limits{\text{h}\rightarrow0}\frac{(\text{x}^2+\text{h}^2+\text{2hx})(\sin\text{x}.\cos\text{x}+\cos\text{x}\sin\text{h})-\text{x}^2\sin\text{x}}{\text{h}}$ $[\because\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{x}^2\sin\text{x}(\cos\text{h}-1)}{\text{h}}+\frac{\text{h}(\text{h}+\text{2x})\sin\text{x}.\cos\text{h}}{\text{h}}+(\text{x}+\text{h})^2\cos\text{x}\frac{\sin\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}-\text{x}^2\sin\text{x}\times\frac{2\sin^2\frac{\text{h}}{2}}{\Big(\frac{\text{h}}{2}\Big)^2}\times\frac{\text{h}^2}{4}+(\text{h}+\text{2x})\sin\text{x}.\cos\text{2h}+(\text{x}+\text{h})^2\cos\text{x}$
$=0+(2\text{x}\sin\text{x}+\text{x}^2\cos\text{x})$
$=\text{2x}\sin\text{x}+\text{x}^2\cos\text{x}$
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Question 45 Marks
Differentiate the following from the first principle$\sqrt{\sin\text{2x}}$
Answer
We 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 $\Big(\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin\text{2x}}\Big)$
$\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sqrt{\sin\text{2x}}}{\text{h}}\times\frac{\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin\text{2x}}}{\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin2\text{x}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h})-\sin\text{2x}}{\Big(\sqrt{\sin(\text{2x}+\text{2h})}+\sqrt{\sin2\text{x}}\Big)}$ $\Big[\sin\text{c}-\sin\text{d}=2\cos\frac{\text{c}+\text{d}}{2}\sin\frac{\text{c}-\text{d}}{2}\Big]$
$\lim_\limits{\text{h}\rightarrow0}\frac{2\cos(\text{2x}+\text{h})\times\sin\text{h}}{\text{h}}\times\frac{1}{\sqrt{\sin(\text{2x}+\text{2h})}+\sqrt{\sin\text{2x}}}$
$=\frac{2\cos\text{2x}}{2\sqrt{\sin\text{2x}}}$
$=\frac{\cos\text{2x}}{\sqrt{\sin\text{2x}}}$
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Question 55 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 $\text{f}'(1)=100\text{f}'(0).$
Answer
We 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)
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Question 65 Marks
Differentiate the following from first principle$\sqrt{\tan\text{x}}$
Answer
We 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 $\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan(\text{x}+\text{h})-\tan{\text{x}}}{\text{h}\big(\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big)}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}+\text{h}-\text{x})}{\text{h}.\cos(\text{x}+\text{h})\cos\text{x}\big[\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big]}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{1}{\cos(\text{x}+\text{h})\cos\text{x}\big[\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big]}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{1}{\cos^2\text{x}.2\sqrt{\tan\text{x}}}\ \Bigg[\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Bigg]$
$=\frac{1}{2}\frac{\sec^2\text{x}}{\sqrt{\tan\text{x}}}\ \Bigg[\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Bigg]$
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Question 75 Marks
Differentiate the following from first principle$\tan\text{2x}$
Answer
$\text{f}(\text{x})=\tan\text{2x}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan2(\text{x}+\text{h})-\tan2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}-\text{2x})}{\text{h}.\cos(\text{2x}+\text{2h})\cos\text{2x}}\ \Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin\text{A}-\text{B}}{\cos\text{A}.\cos{\text{B}}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{\text{h}.\cos(\text{2x}+\text{2h})\cos2\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin2\text{h}}{2\text{h}}\Big)\times\frac{1\times2}{\cos(\text{2h}+\text{2x})\cos2\text{x}}$
$=\frac{2}{\cos2\text{x}.\cos2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2\text{h}}=1\Big]$
$=2\sec^22\text{x}\ \Big[\because\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Big]$
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Question 85 Marks
Differentiate the following from first principle$\tan\text{2x}$
Answer
$\text{f}(\text{x})=\tan\text{2x}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan2(\text{x}+\text{h})-\tan2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}-\text{2x})}{\text{h}.\cos(\text{2x}+\text{2h})\cos\text{2x}}\ \Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin\text{A}-\text{B}}{\cos\text{A}.\cos{\text{B}}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{\text{h}.\cos(\text{2x}+\text{2h})\cos2\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin2\text{h}}{2\text{h}}\Big)\times\frac{1\times2}{\cos(\text{2h}+\text{2x})\cos2\text{x}}$
$=\frac{2}{\cos2\text{x}.\cos2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2\text{h}}=1\Big]$
$=2\sec^22\text{x}\ \Big[\because\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Big]$
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Question 95 Marks
Differentiate the following from first principle$\tan\text{x}^2$
Answer
We have,$\text{f}(\text{x})=\tan\text{x}^2$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan{(\text{x}+\text{h})^2}-\tan{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2}{\cos(\text{x}+\text{h})}-\frac{\sin\text{x}^2}{\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2\cos\text{x}^2-\cos(\text{x}+\text{h})^2\sin\text{x}^2}{\cos(\text{x}+\text{h})^2\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big((\text{x}+\text{h})^2-\text{x}^2\Big)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}^2+\text{h}^2+\text{2xh}-\text{x}^2)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h}^2+\text{2xh})}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2 }$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{(\text{h}+\text{2x})}{\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=1.\frac{\text{2x}}{\cos^2(\text{x})^2}$
$=\text{2x}\sec^2\text{x}^2$
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Question 105 Marks
Differentiate the following from the first principle$\cos\Big(\text{x}-\frac{\pi}{8}\Big)$
Answer
Let $\text{f}(\text{x})=\cos\Big(\text{x}-\frac{\pi}{8}\Big).$ 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)$
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Question 115 Marks
Differentiate the following from first principle$\tan\text{x}^2$
Answer
We 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$
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Question 125 Marks
Differentiate the following from first principle$\cos\sqrt{\text{x}}$
Answer
We 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)$
$=\lim_\limits{\text{h}\rightarrow0}-1\frac{\text{h}}{\text{h}(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})}\times\sin\Bigg(\frac{\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}}{2}\Bigg)$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-1}{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}\sin\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{2}$
$=\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
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Question 135 Marks
Differentiate the following from first principle$\tan^2\text{x} $
Answer
$\text{f}(\text{x})=\tan^2\text{x}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan^2(\text{x}+\text{h})-\tan^2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\big\{\tan(\text{x}+\text{h})+\tan{\text{x}}\big\}\big\{\tan(\text{x}+\text{h})-\tan{\text{x}}\big\}}{\text{h}}$
$\big[\because\tan^2\text{A}-\tan^2\text{B}=(\tan\text{A}+\tan\text{B})(\tan\text{A}-\tan\text{B})\big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h}+\text{x})}{\cos(\text{x}+\text{h})\cos\text{x}}\times\frac{\sin(\text{x}+\text{h}-\text{x})}{\cos(\text{x}+\text{h})\cos\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{h})}{\text{h}.\cos(\text{x}+\text{h})\cos\text{x}}\times\frac{\sin\text{h}}{\cos(\text{x}+\text{h})\cos\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{\sin2\text{x}}{\cos^2\text{x}.\cos^2(\text{x}+\text{h})}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{x}}{\cos^2\text{x}.\cos^2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\text{x}.\cos\text{x}}{\cos^2\text{x}}\times\frac{1}{\cos^2\text{x}}\ [\sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\lim_\limits{\text{h}\rightarrow0}2\tan\text{x}.\sec^2\text{x}$
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Question 145 Marks
Differentiate the following from first principle:$-\text{x}$
Answer
Let $\text{f}\text{(x)}=-\text{x.}$ Then, $\text{f}(\text{x}+\text{h})=(-\text{x}+\text{h})$$\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}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-(\text{x}+\text{h})+(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}-1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=-1$
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Question 155 Marks
Differentiate the following from first principle$\tan\sqrt{\text{x}}$
Answer
We have,$\text{f}(\text{x})=\tan\sqrt{\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan\sqrt{(\text{x}+\text{h})}-\tan\sqrt{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}.\cos\sqrt{\text{x}+\text{h}}\cos\sqrt{\text{x}}} \ \Bigg[\because\tan\text{A}-\tan\text{B}=\frac{\sin(\text{A}-\text{B})}{\cos\text{A}.\cos\text{B}}\Bigg]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\text{x}+\text{h}-\text{x})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}\times\frac{1}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=1\times\frac{1}{2\sqrt{\text{x}}\cos\sqrt{\text{x}}\cos\sqrt{\text{x}+\text{h}}}\ \Bigg[\because\lim_\limits{\text{h}\rightarrow0}=\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}=1\Bigg]$
$=\frac{1}{2\sqrt{\text{x}}\cos^2\text{x}}$
$=\frac{\sec^2\text{x}}{2\sqrt{\text{x}}}$
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Question 165 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.
Answer
Let $\text{u}=1+2\tan\text{x};\text{v}=5+4\cos\text{x}$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,
$\frac{\text{d}}{\text{dx}}[(1+2\tan\text{x})(5+4\cos\text{x})]\\=\frac{\text{d}}{\text{dx}}(5+4\cos\text{x}+10\tan\text{x}+8\sin\text{x})$
$=-4\sin\text{x}+10\sec^2\text{x}+8\cos\text{x}$
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