Question

If $y={\cos }^{-1}\left(\sqrt{x}\right)$, then find $\frac{dy}{dx}$ using first principle.

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

Answer 1

The correct option is C $-\frac{1}{2\sqrt{x}\sqrt{1-x}}$

Given, $f\left(x\right)={\cos }^{-1}\left(\sqrt{x}\right)$

Thus $f(x+h)={\cos }^{-1}\left(\sqrt{x+h}\right)$

We know that $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{{\cos }^{-1}\left(\sqrt{x+h}\right)-{\cos }^{-1}\left(\sqrt{x}\right)}{h}$

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(\frac{\pi }{2}-{\sin }^{-1}(\sqrt{x+h}\left)\right)-(\frac{\pi }{2}-{\sin }^{-1}(\sqrt{x}\left)\right)}{h}$

$f^{\prime }\left(x\right)=-{lim}_{h\to 0}\frac{{\sin }^{-1}\left(\sqrt{x+h}\right)-{\sin }^{-1}\left(\sqrt{x}\right)}{h}=-{lim}_{h\to 0}\frac{{\sin }^{-1}\left(\sqrt{x+h}\right(\sqrt{1-(\sqrt{x}{)}^2})-\sqrt{x}(\sqrt{1-(\sqrt{x+h}{)}^2})}{h}$

Multiply and divide by $\sqrt{x+h}\sqrt{1-(\sqrt{x}{)}^2}-\sqrt{x}\sqrt{1-(\sqrt{x+h}{)}^2}$

$=-{lim}_{h\to 0}\frac{\sqrt{x+h}\sqrt{1-(\sqrt{x}{)}^2}-\sqrt{x}\sqrt{1-(\sqrt{x+h}{)}^2}}{h}$

Multiply and divide by $\sqrt{x+h}\sqrt{1-(\sqrt{x}{)}^2}+\sqrt{x}\sqrt{1-(\sqrt{x+h}{)}^2}$

$=-{lim}_{h\to 0}\frac{(\sqrt{x+h}{)}^2-(\sqrt{x}{)}^2}{h(\sqrt{x+h}\sqrt{1-(\sqrt{x}{)}^2}+\sqrt{x}\sqrt{1-(\sqrt{x+h}{)}^2})}=-\frac{1}{2\sqrt{x}\sqrt{1-x}}$