Question

FIND THE FIRST DERIVATIVE OF g(x)=√x² + 2x​​

Answer 1

We are asked to find the first derivative of,

$\longrightarrow g(x) = \sqrt{x^2+2x}$

Here we can write the term $x^2+2x$ by adding and subtracting 1 as,

$x^2+2x = x^2+2x+1-1$

$x^2+2x = (x+1)^2-1\quad[\because\, x^2+2x+1=(x+1)^2]$

Thus,

$\longrightarrow g(x) = \sqrt{(x+1)^2-1}\quad\dots(1)$

Now take,

$x+1=\sec\theta\quad\dots(2)$

$x=\sec\theta-1$

$dx=\sec\theta\tan\theta\, d\theta$

$\dfrac{d\theta}{dx}=\dfrac{1}{\sec\theta\tan\theta}\quad\dots(3)$

Then (1) becomes,

$\longrightarrow g(x) = \sqrt{sec^2\theta-1}$

We have,

$\sec^2\theta-1=\tan^2\theta$

So we get,

$\longrightarrow g(x) = \tan\theta$

Now,

$\longrightarrow g'(x) = \dfrac{d}{dx}\,[\tan\theta]$

By chain rule,

$\longrightarrow g'(x) = \dfrac{d}{d\theta}\,[\tan\theta]\cdot\dfrac{d\theta}{dx}$

$\longrightarrow g'(x) = \sec^2\theta\cdot\dfrac{1}{\sec\theta\tan\theta}\quad\quad\textrm{[From (3)]}$

$\longrightarrow g'(x) = \sec\theta\cdot\dfrac{1}{\tan\theta}$

$\longrightarrow g'(x)=\dfrac{1}{\cos\theta}\cdot\dfrac{\cos\theta}{\sin\theta}$

$\longrightarrow g'(x)=\dfrac{1}{\sin\theta}\quad\dots(4)$

But we have,

$\sin^2\theta+\cos^2\theta=1$

$\sin\theta=\sqrt{1-\cos^2\theta}$

$\sin\theta=\sqrt{1-\dfrac{1}{\sec^2\theta}}$

$\sin\theta=\dfrac{\sqrt{\sec^2\theta-1}}{\sec\theta}$

$\sin\theta=\dfrac{\sqrt{(x+1)^2-1}}{x+1}\quad\quad\textrm{[From (2)]}$

$\sin\theta=\dfrac{\sqrt{x^2+2x}}{x+1}$

Hence (4) becomes,

$\longrightarrow\underline{\underline{g'(x)=\dfrac{x+1}{\sqrt{x^2+2x}}}}$

This is the first derivative of the given function.