How do you differentiate #y=1/4arctan(x/4)#?

1 Answer
Jim G.
Jan 11, 2018

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

Explanation:

#•color(white)(x)d/dx(arctanx)=1/(1+x^2)#

#"differentiate using the "color(blue)"chain rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#y=1/4arctan(x/4)#

#rArrdy/dx=1/4xx1/(1+(x/4)^2)xxd/dx(x/4)#

#color(white)(rArrdy/dx)=1/4xx1/(1+(x^2/16))xx1/4#

#color(white)(rArrdy/dx)=1/16xx1/(1+(x^2/16))=1/(16+x^2)#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
1757 views around the world
You can reuse this answer
Creative Commons License