Differentiate each of the following functions from the first principal :

Solution:

We have to find the derivative of $\sin ^{-1}(2 x+3)$ with the first principle method, so,

$f(x)=\sin ^{-1}(2 x+3)$

by using the first principle formula, we get,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sin ^{-1}(2[x+h]+3)-\sin ^{-1}(2 x+3)}{h}$

Let $\sin ^{-1}[2(x+h)+3]=A$ and $\sin ^{-1}(2 x+3)=B$, so,

$\sin A=[2(x+h)+3]$ and $\sin B=(2 x+3)$

$2 h=\sin A-\sin B$, when $h \rightarrow 0$ then $\sin A \rightarrow \sin B$ we can also say that $A \rightarrow B$ and hence $A-B \rightarrow 0$,

$f^{\prime}(x)=\lim _{A-B \rightarrow 0} \frac{2(A-B)}{\sin A-\sin B}$

$f^{\prime}(x)=\lim _{A-B \rightarrow 0} \frac{2(A-B)}{2 \sin \frac{A-B}{2} \cos \frac{A+B}{2}}$

$\left[\sin C-\sin D=2 \sin \frac{C-D}{2} \cos \frac{C+D}{2}\right]$

$f^{\prime}(x)=\lim _{A-B \rightarrow 0} \frac{2}{1 \cos \frac{A+B}{2}}$

[By using $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ ]

$f^{\prime}(x)=\frac{2}{\cos B}$

$f^{\prime}(x)=\frac{2}{\cos \left[\sin ^{-1}(2 x+3)\right]}$

[By using Pythagoras theorem, in which $\mathrm{H}=1$ and $\mathrm{P}=2 \mathrm{x}+3$, so, we have to find $\mathrm{B}$, which comes out to be $\sqrt{1-(2 x+3)^{2}}$ by the relation $\left.H^{2}=P^{2}+B^{2}\right]$

$f^{\prime}(x)=\frac{2}{\sqrt{1-(2 x+3)^{2}}}$