# Derivative of 1/sqrt(1-x^2) | Derivative of 1/sqrt(a^2-x^2)

## Derivative of  $1/\sqrt{1-x^2}$

First Method: At first, we will find the derivative of 1/root(1-$x^2$) by the quotient rule of derivatives. Let us recall the quotient rule of derivatives. If $f$ and $g$ be two functions then the derivative of $f/g$ is given by the following formula:

$\dfrac{d}{dx}(f/g)$ $=\dfrac{gf’ -f g’}{g^2}$ $\cdots (\star)$

Here $’$ denotes the derivative with respect to $x.$

By the above rule, the derivative of $1/\sqrt{1-x^2}$ is equal to

$\dfrac{d}{dx}(1/\sqrt{1-x^2})$ $=\dfrac{\sqrt{1-x^2}\cdot 1′ – 1\cdot (\sqrt{1-x^2})’ }{(\sqrt{1-x^2})^2}$

As $\sqrt{t}=t^{1/2}$ and the power rule of derivatives $d/dt(t^n)=nt^{n-1}$, we have that

$\dfrac{d}{dx}(1/\sqrt{1-x^2})=$ $\dfrac{\sqrt{1-x^2}\cdot 0 -1/2(1-x^2)^{-1/2}\cdot -2x}{1-x^2}$

$=\dfrac{0+x(1-x^2)^{-1/2}}{1-x^2}$

$=\dfrac{x}{(1-x^2)\sqrt{1-x^2}}$ as we know that $a^{-1/2}=1/\sqrt{a}$

$=\dfrac{x}{(1-x^2)^{3/2}}$

$=\dfrac{x}{(1-x^2)\sqrt{1-x^2}}$

So the derivative of 1/root(1-$x^2$) is equal to $\dfrac{x}{(1-x^2)^{3/2}}.$

Second Method: Now, we will evaluate the derivative of 1/root(1-$x^2$) by the substitution method and the chain rule of derivatives.

Let $z=1-x^2$ $\Rightarrow \dfrac{dz}{dx} = -2x$

$\therefore \dfrac{d}{dx}(1/\sqrt{1-x^2})$ $=\dfrac{d}{dx}(1/\sqrt{z})$

$\Rightarrow \dfrac{d}{dx}(1/\sqrt{1-x^2})$  $=\dfrac{d}{dz}(1/\sqrt{z}) \cdot \dfrac{dz}{dx}$ by the chain rule of derivatives.

$\Rightarrow \dfrac{d}{dx}(1/\sqrt{1-x^2})$ $=\dfrac{d}{dz}(z^{-1/2}) \cdot (-2x)$

$=-1/2 z^{-1/2-1} \cdot (-2x)$ as we know that $\dfrac{d}{dx}(x^n)=nx^{n-1}$ (power rule of derivatives)

$=xz^{-3/2}$

$=\dfrac{x}{(1-x^2)^{3/2}}$ as $z=1-x^2.$

## Derivative of  $1/\sqrt{a^2-x^2}$

We will calculate the derivative of $1/\sqrt{a^2-x^2}$ by the chain rule of derivatives.

Let $z=a^2-x^2$ $\Rightarrow \dfrac{dz}{dx} = -2x$

$\therefore \dfrac{d}{dx}(1/\sqrt{a^2-x^2})$ $=\dfrac{d}{dx}(1/\sqrt{z})$

$\Rightarrow \dfrac{d}{dx}(1/\sqrt{a^2-x^2})$  $=\dfrac{d}{dz}(1/\sqrt{z}) \cdot \dfrac{dz}{dx}$ by the chain rule of derivatives.

$\Rightarrow \dfrac{d}{dx}(1/\sqrt{a^2-x^2})$ $=\dfrac{d}{dz}(z^{-1/2}) \cdot (-2x)$

$=-1/2 z^{-1/2-1} \cdot (-2x)$ by the power rule of derivatives

$=xz^{-3/2}$

$=\dfrac{x}{(a^2-x^2)^{3/2}}$ as $z=a^2-x^2.$

$=\dfrac{x}{(a^2-x^2)\sqrt{a^2-x^2}}$

Thus the derivative of 1/root($a^2-x^2$) is $\dfrac{x}{(a^2-x^2)\sqrt{a^2-x^2}}.$

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