`f(x) = (x^3 + 4x)(3x^2 + 2x - 5), c=0` Find f'(x) and f'(c).

Expert Answers
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You need to first find derivative of the function using the product rule:

`f'(x)= (x^3 + 4x)'*(3x^2+2x-5)+(x^3 + 4x)*(3x^2+2x-5)'`

`f'(x)= (3x^2 + 4)*(3x^2+2x-5)+(x^3 + 4x)*(6x+2)` 

`f'(x)= 9x^4 + 6x^3 - 15x^2 + 12x^2 + 8x - 20 + 6x^4 + 2x^3 + 24x^2 +8x`

Combining like terms yields:

`f'(x)=15x^4 + 8x^3 + 21x^2 + 16x - 20`

Hence, evaluating the derivative of the function, yields` f'(x)=15x^4 + 8x^3 + 21x^2 + 16x - 20.`

You need to evaluate f'(c) at c = 0, hence, you need to replace 0 for x in equation of f'(x).

`f'(0)=15*0^4 + 8*0^3 + 21*0^2 + 16*0 - 20` 

`f'(0)=-20`

Hence, evaluating the derivative of the function, yields `f'(x)=15x^4 + 8x^3 + 21x^2 + 16x - 20` and evaluating f'(c) at c = 0, yields f'(0)= -20.