Question

If the function $f\left(x\right)=\sqrt{1-x^2}$, then find $f^{\prime }\left(x\right)$.

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

Answer 1

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

We have,

$f\left(x\right)=\sqrt{1-x^2}$

On differentiating w.r.t $x$, we get

$f^{\prime }\left(x\right)=\frac{d}{dx}\left(\sqrt{1-x^2}\right)$

$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{1-x^2}}\times (0-2x)$

$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{1-x^2}}\times (-2x)$

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

Hence, this is the answer.