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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions from first principles:
$\sin^{-1}(2\text{x}+3)$
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Answer

Let $\text{f(x)}=\sin^{-1}(2\text{x}+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2(\text{x}+\text{h})+3)$
$\Rightarrow\ \text{f}(\text{x}+\text{h})=\sin^{-1}(2\text{x}+2\text{h}+3)$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}(2\text{x}+2\text{h}+3)-\sin^{-1}(2\text{x}+3)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\Big[(2\text{x}+2\text{h}+3)\sqrt{1+(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big]}{\text{h}}$
$\Big[\text{Since}, \sin^{-1}\text{x}-\sin^{-1}\text{y}=\sin^{-1}\big[\text{x}\sqrt{1-\text{y}^2}-\text{y}\sqrt{1-\text{x}^2}\big]\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{z}}{\text{z}}\times\frac{\text{z}}{\text{h}}$
Where, $\text{z}=(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}$
$\text{and }\lim\limits_{\text{h}\rightarrow0}\frac{\sin^{-1}\text{h}}{\text{h}}=1$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{z}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{x}+2\text{h}+3)^2-(2\text{x}+3)^2-(2\text{x}+3)^2\big(1-(2\text{x}+2\text{h}+3)^2\big)}{\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}-(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
[Since, rationalizing numerator]
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)\big]\big(1-(2\text{x}+3)^2\big)-(2\text{x}+3)^2\big[1-(2\text{x}+3)^2-4\text{h}^2-4\text{h}(2\text{x}+3)\big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\Big[(2\text{x}+3)^2+4\text{h}^2+4\text{h}(2\text{x}+3)-(2\text{x}+3)^4-4\text{h}^2(2\text{x}+3)^2-4\text{h}(2\text{x}+3)^3 -(2\text{x}+3)^2+(2\text{x}+3)^3+4\text{h}^2(2\text{x}+3)^2+4\text{h}(2\text{x}+3)^3\Big]}{\text{h}\big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\big\}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{\text{h}\Big\{(2\text{x}+2\text{h}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+2\text{h}+3)^2}\Big\}}$
$=\frac{4\text{h}\big[\text{h}+(2\text{x}+3)\big]}{(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}+(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{4(2\text{x}+3)}{2(2\text{x}+3)\sqrt{1-(2\text{x}+3)^2}}$
$=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$
So,
$\frac{\text{d}}{\text{dx}}\big(\sin^{-1}(2\text{x}+3)\big)=\frac{2}{\sqrt{1-(2\text{x}+3)^2}}$

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