Question

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

Answer 1

We are asked to find the first derivative of,

`\longrightarrow g(x) = √(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) = √((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`

`(d\theta)/(dx)=(1)/(\sec\theta\tan\theta)\quad\dots(3)`

Then (1) becomes,

`\longrightarrow g(x) = √(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) = (d)/(dx)\,[\tan\theta]`

By chain rule,

`\longrightarrow g'(x) = (d)/(d\theta)\,[\tan\theta]\cdot(d\theta)/(dx)`

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

`\longrightarrow g'(x) = \sec\theta\cdot(1)/(\tan\theta)`

`\longrightarrow g'(x)=(1)/(\cos\theta)\cdot(\cos\theta)/(\sin\theta)`

`\longrightarrow g'(x)=(1)/(\sin\theta)\quad\dots(4)`

But we have,

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

`\sin\theta=√(1-\cos^2\theta)`

`\sin\theta=\sqrt{1-(1)/(\sec^2\theta)}`

`\sin\theta=(√(\sec^2\theta-1))/(\sec\theta)`

`\sin\theta=(√((x+1)^2-1))/(x+1)\quad\quad\textrm{[From (2)]}`

`\sin\theta=(√(x^2+2x))/(x+1)`

Hence (4) becomes,

`\longrightarrow\underline{\underline{g'(x)=(x+1)/(√(x^2+2x))}}`

This is the first derivative of the given function.