What is the derivative of # (x^3+1)^2#?

3 Answers
sankarankalyanam
Mar 31, 2018

#color(indigo)(f'(x) = 6x^2 * (x^3 + 1)#

Explanation:

#f(x) = (x^3 + 1)^2#

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Applying the chain rule,

#f'(x) = 2 * (x^3 + 1) * (d/(dx)) (x^3 + 1)#

#f'(x) = 2 * (x^3 + 1) * 3x^2#

#color(indigo)(f'(x) = 6x^2 * (x^3 + 1)#

Jim G.
Mar 31, 2018

#6x^5+6x^2#

Explanation:

#"expand the factor "(x^3+1)^2#

#rArr(x^3+1)^2=x^6+2x^3+1#

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

#•color(white)(x)d/dx(ax^n)=nax^(n-1)#

#rArrd/dx((x^3+1)^2)=6x^5+6x^2#

Brandon
Mar 31, 2018

#6x^5+6x^2#

Explanation:

Firstly, expand the brackets

#(x^3+1)(x^3+1)=x^6+x^3+x^3+1=x^6+2x^3+1#

Now take the derivative using the power rule:

#d/dx# #(x^6+2x^3+1)#

#-> d/dx=6x^5+6x^2#

Remembering when deriving, multiply the power by the coefficient, and then #-1# from the power.

Also remembering that if there is no #x#, it cannot be derived

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