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How do you differentiate #f(x)=(3x^3-2x^2+5)^331#?

1 Answer

Oct 11, 2016

#(dy)/(dx)=331(9x^2-4x)(3x^3-2x^2+5)^330#

Explanation:

Using chain rule: #(dy)/(dx)=(dy)/(du)*(du)/(dx)#

In this case, #y=(3x^3-2x^2+5)^331#

Let #u=3x^3-2x^2+5#,

then #(dy)/(du)=331u^330# and #(du)/(dx)=9x^2-4x#

So #(dy)/(dx)=331u^330*(9x^2-4x)#

#=331(9x^2-4x)(3x^3-2x^2+5)^330#

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