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

2 Answers
Jun 7, 2016

#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))#

Jun 7, 2016

#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)}#