What is the derivative of this function #1/4arctan(x/4)#?

1 Answer
mason m
Dec 6, 2016

#1/(x^2+16)#

Explanation:

It's helpful to know that #d/dxarctan(x)=1/(x^2+1)#. Then, when there is a function within the arctangent function, we can apply the chain rule to see that #d/dxarctan(f(x))=1/((f(x))^2+1)*f'(x)#.

Thus, #d/dx1/4arctan(x/4)=1/4*1/((x/4)^2+1)*d/dx(x/4)#

Continuing simplification, #d/dx1/4arctan(x/4)=1/4*1/(x^2/16+1)*1/4#

#=1/16*1/((x^2+16)/16)#

#=1/16*16/(x^2+16)#

#=1/(x^2+16)#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
1993 views around the world
You can reuse this answer
Creative Commons License