How do you find f'(0) given #f(x) =( (x^5+5)/(3cosx))^2#?

1 Answer
Jim H
Dec 22, 2016

#f'(0) = 2((0^5+5)/(3cos0)) *0 = 0# See below.

Explanation:

#f'(x) = 2((x^5+5)/(3cosx)) * d/dx((x^5+5)/(3cosx))#

Notice that when we apply the quotient rule to find

#d/dx((x^5+5)/(3cosx))#,

The derivatives of the numerator (#5x^4#) and denominator #(-3sinx)# are both #0# at #x=0#, so lets not bother with it.

#f'(0) = 2((0^5+5)/(3cos0)) *0 = 0#

Related questions
See all questions in Chain Rule
Impact of this question
1131 views around the world
You can reuse this answer
Creative Commons License