How do you use the chain rule to differentiate #f(x)=sin(x^2)/(x^4-3x)^4#?

1 Answer
Andrew B.
Nov 5, 2017

#f'(x)=((2x^5-6x^2)cos(x^2)-(16x^2+12)sin(x^2))/(x^4-3x)^5#

Explanation:

This is a rational expression that will require us to use the quotient rule to differentiate. First let's rewrite the function as a product since the product rule is easier to use.

#f(x)=(sin(x^2))/(x^4-3x)^4# #= sin(x^2)(x^4-3x)^-4#

Recall that for a function of the form #f(x)=a*b#

The product rule for differentiation is as follows:
#f'(x)=a'b+ab'#

Also notice that in our function, #a# and #b# are both compositions. Recall that the derivative of a composition requires us to use the chain rule.

For a function of the form #f(x)=f(g(x))#, the derivative is #f'(x)=f'(g(x))*g'(x)#

So, applying the product rule and chain rule on our function, we have
#f'(x)=[cos(x^2)*2x]##[(x^4-3x)^-4]#+#[sin(x^2)][-4(x^4-3x)^-5# #(4x^3-3)]#

#f'(x)=(2xcos(x^2))/(x^4-3x)^4#- #(4(4x^2-3)sin(x^2))/(x^4-3x)^5#

#f'(x)= (x^4-3x)/(x^4-3x)##(2xcos(x^2))/(x^4-3x)^4#- #(4(4x^2-3)sin(x^2))/(x^4-3x)^5#

#f'(x)=((x^4-3x)2xcos(x^2)-4(4x^2-3)sin(x^2))/(x^4-3x)^5#

#f'(x)=((2x^5-6x^2)cos(x^2)-(16x^2+12)sin(x^2))/(x^4-3x)^5#

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