Question

What is the derivative of `y=tan^-1 sqrt(3x)`?

Answer 1

`dy/dx = 3/(2(1+3x)sqrt(3x))`

Explanation:

Apply the chain rule in a straightforward manner.

`dy/dx = 1/(1 + (sqrt(3x))^2) times 1/(2sqrt(3x)) times 3 = 3/(2(1+3x)sqrt(3x))`

Answer 2

`frac{d}{dx}(arctan (sqrt{3x}))=frac{sqrt{3}}{2sqrt{x}(3x+1)}`

Explanation:

`frac{d}{dx}(arctan (sqrt{3x}))`

Applying chain rule,
`frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}`
Let ,`sqrt(3x)`=`u`
`=frac{d}{du}(arctan (ut))frac{d}{dx}(sqrt{3x})`
`frac{d}{du}(arctan (u))=frac{1}{u^2+1}`
`frac{d}{dx}(sqrt{3x})=frac{sqrt{3}}{2sqrt{x}}`
so,
`=frac{1}{u^2+1}frac{sqrt{3}}{2sqrt{x}}`
substitute back,`u=\sqrt{3x}`
`=frac{1}{(sqrt{3x})^2+1}frac{sqrt{3}}{2sqrt{x}}`

Simplifying back,
`=frac{sqrt{3}}{2sqrt{x}(3x+1)}`