Question

Find derivative of ${\sin }^{-1}\left(x^2\right)$ using first principle.

• A. $\frac{2x}{\sqrt{1-x^2}}$
• B. $\frac{2x}{\sqrt{1-x}}$
• C. $\frac{2x}{\sqrt{1-x^4}}$
• D. $\frac{x}{\sqrt{1-x^4}}$

Answer 1

Answer: (C)

$\frac{2x}{\sqrt{1-x^4}}$

It is given to us that ${\sin }^{-1}\left(x^2\right)$ = y, we can write $x^2$ = $\sin \left(y\right)$, so x = $\sqrt{\sin \left(y\right)}$,

Now $\frac{dx}{dy}$ = $li{m}_{h\to 0}\frac{\sqrt{\sin (y+h)}-\sqrt{\sin \left(y\right)}}{h}$ (by using the first principle)

Now Rationalising the numerator by multiplying and dividing it by: ($\sqrt{\sin (y+h)}+\sqrt{\sin \left(y\right)}$ )

After rationalisation we get: $li{m}_{h\to 0}\frac{\sin (y+h)-\sin \left(y\right)}{h(\sqrt{\sin (y+h)}+\sqrt{\sin \left(y\right)})}$

Now using the formula: $\sin \left(A\right)-\sin \left(B\right)=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

We can write:-

$li{m}_{h\to 0}\frac{2\cos \left(y\right)\sin \left(\frac{h}{2}\right)}{h(\sqrt{\sin (y+h)}+\sqrt{\sin \left(y\right)}}$ = $li{m}_{h\to 0}\frac{\cos \left(y\right)}{\sqrt{\sin (y+h)}+\sqrt{\sin \left(y\right)}}$$\frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$

Using the formula: $li{m}_{h\to 0}\frac{\sin \left(h\right)}{h}$ = 1, also $\sin (y+h)$ is equal to $\sin \left(y\right)$ when h is very very small

so after applying the limits we get $\frac{dx}{dy}$ = $\frac{\cos \left(y\right)}{2\sqrt{\sin \left(y\right)}}$

using ${\sin }^2\left(a\right)+{\cos }^2\left(a\right)=1$, we can write $\cos \left(a\right)=\sqrt{1-{\sin }^2\left(a\right)}$ and from the begining we can see that$\sqrt{\sin \left(y\right)}$ = x

So finally we get $\frac{dx}{dy}$ = $\frac{\sqrt{1-x^4}}{2x}$

But we want the value of $\frac{dy}{dx}$ which will be $\frac{2x}{\sqrt{1-x^4}}$