How do you differentiate #f(x)=(x^3-x)(x+3) # using the product rule?

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Answer 1 · 2016-01-02T22:24:46

#f'(x)=4x^3+9x^2-3#

#d/dx[f(x)g(x)]=f'(x)g(x)+g'(x)f(x)#

Thus,

#f'(x)=(x+3)d/dx(x^3-x)+(x^3-x)d/dx(x+3)#

Find each derivative.

#d/dx(x^3-x)=3x^2-1#

#d/dx(x+3)=1#

Plug these both back in.

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

Simplify.

#f'(x)=4x^3+9x^2-3#