How do you use the chain rule to differentiate #y=(5x^4+1)^2#?

1 Answer
Jim G.
May 1, 2017

#dy/dx=200x^7+40x^3#

Explanation:

#d/dx(f(g(x)))=f'(g(x))xxg'(x)larr" chain rule"#

#"here "f(g(x))=(5x^4+1)^2#

#g(x)=5x^4+1#

#y=(5x^4+1)^2#

#rArrdy/dx=2(5x^4+1)xxd/dx(5x^4+1)#

#color(white)(rArrdy/dx)=2(5x^4+1)xx20x^3#

#color(white)(rArrdy/dx)=40x^3(5x^4+1)#

#color(white)(rArrdy/dx)=200x^7+40x^3#

Related questions
See all questions in Chain Rule
Impact of this question
3398 views around the world
You can reuse this answer
Creative Commons License