5 Marks Questions

Question 515 Marks

Differentiate the following from first principle$\tan\text{x}^2$

Answer

We have,$\text{f}(\text{x})=\tan\text{x}^2$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan{(\text{x}+\text{h})^2}-\tan{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2}{\cos(\text{x}+\text{h})}-\frac{\sin\text{x}^2}{\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2\cos\text{x}^2-\cos(\text{x}+\text{h})^2\sin\text{x}^2}{\cos(\text{x}+\text{h})^2\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big((\text{x}+\text{h})^2-\text{x}^2\Big)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}^2+\text{h}^2+\text{2xh}-\text{x}^2)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h}^2+\text{2xh})}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2 }$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{(\text{h}+\text{2x})}{\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=1.\frac{\text{2x}}{\cos^2(\text{x})^2}$
$=\text{2x}\sec^2\text{x}^2$

View full question & answer→

Question 525 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}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Bigg[\frac{\big(\text{x}+\text{h}-\frac{\pi}{8}\big)+\big(\text{x}-\frac{\pi}{8}\big)}{2}\Bigg]\sin\Bigg[\frac{\big(\text{x}+\text{h}-\frac{\pi}{8}\big)-\big(\text{x}-\frac{\pi}{8}\big)}{2}\Bigg]}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{-2\sin\Bigg[\frac{\text{2x}+\text{h}-\frac{2\pi}{8}}{2}\Bigg]\sin\Big[\frac{\text{h}}{2}\Big]}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}-\sin\Bigg[\frac{2\text{x}+\text{h}-\frac{2\pi}{8}}{2}\Bigg]\times\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big[\frac{\text{h}}{2}\Big]}{\frac{\text{h}}{2}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=-\sin\Bigg[\frac{\text{2x}+0-\frac{2\pi}{8}}{2}\Bigg]\times1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=-\sin\Big(\text{x}-\frac{\pi}{8}\Big)$

View full question & answer→

Question 535 Marks

Differentiate the following from first principle$\tan\text{x}^2$

Answer

We have,$\text{f}(\text{x})=\tan\text{x}^2$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan{(\text{x}+\text{h})^2}-\tan{\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2}{\cos(\text{x}+\text{h})}-\frac{\sin\text{x}^2}{\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{\sin(\text{x}+\text{h})^2\cos\text{x}^2-\cos(\text{x}+\text{h})^2\sin\text{x}^2}{\cos(\text{x}+\text{h})^2\cos\text{x}^2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big((\text{x}+\text{h})^2-\text{x}^2\Big)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}^2+\text{h}^2+\text{2xh}-\text{x}^2)}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h}^2+\text{2xh})}{\text{h}.\cos(\text{x}+\text{h})^2.\cos\text{x}^2 }$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{(\text{h}+\text{2x})}{\cos(\text{x}+\text{h})^2.\cos\text{x}^2}$
$=1.\frac{\text{2x}}{\cos^2(\text{x})^2}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$
$=\text{2x}\sec^2\text{x}^2$

View full question & answer→

Question 545 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)$
$=\lim_\limits{\text{h}\rightarrow0}-1\frac{\text{h}}{\text{h}(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})}\times\sin\Bigg(\frac{\sqrt{\text{x}+\text{h}}+\sqrt{\text{x}}}{2}\Bigg)$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-1}{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}\sin\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{2}$
$=\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$

View full question & answer→

Question 555 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 565 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}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-(\text{x}+\text{h})+(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}-1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=-1$

View full question & answer→

Question 575 Marks

Differentiate the following from first principle$\tan\sqrt{\text{x}}$

Answer

We have,$\text{f}(\text{x})=\tan\sqrt{\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan\sqrt{(\text{x}+\text{h})}-\tan\sqrt{\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}.\cos\sqrt{\text{x}+\text{h}}\cos\sqrt{\text{x}}} \ \Bigg[\because\tan\text{A}-\tan\text{B}=\frac{\sin(\text{A}-\text{B})}{\cos\text{A}.\cos\text{B}}\Bigg]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\text{x}+\text{h}-\text{x})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})(\sqrt{(\text{x}+\text{h})}+\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}\times\frac{1}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})\cos\sqrt{\text{x}}.\cos\sqrt{\text{x}+\text{h}}}$
$=1\times\frac{1}{2\sqrt{\text{x}}\cos\sqrt{\text{x}}\cos\sqrt{\text{x}+\text{h}}}\ \Bigg[\because\lim_\limits{\text{h}\rightarrow0}=\frac{\sin(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}{(\sqrt{(\text{x}+\text{h})}-\sqrt{\text{x}})}=1\Bigg]$
$=\frac{1}{2\sqrt{\text{x}}\cos^2\text{x}}$
$=\frac{\sec^2\text{x}}{2\sqrt{\text{x}}}$

View full question & answer→

Question 585 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,
$\frac{\text{d}}{\text{dx}}[(1+2\tan\text{x})(5+4\cos\text{x})]\\=\frac{\text{d}}{\text{dx}}(5+4\cos\text{x}+10\tan\text{x}+8\sin\text{x})$
$=-4\sin\text{x}+10\sec^2\text{x}+8\cos\text{x}$

View full question & answer→

MATHS 5 Marks Questions