# Derivative Help for Calculus

Determine the derivative for the following function (Do not simplify your answer):$f(x)= ( x^2+3x−5)^7 (5x^3+4x^2−3x+8)$. So far I got $(2x+3x-5)^7 (15x^2+8x-3x+8)$ What would I do after this step?

Hint:

Use the product rule:

$f'(x)=\frac{d}{dx}(x^2+3x−5)^7\cdot(5x^3+4x^2−3x+8)+(x^2+3x−5)^7\cdot\frac{d}{dx}(5x^3+4x^2−3x+8)$

Can you take it from here?

Help

$\frac{d}{dx}(x^2+3x-5)^7=$

$\frac{d}{dx}(5x^3+4x^2−3x+8)=$

Solution:

$f'(x)=7(x^2+3x-5)^6\cdot(2x+3)\cdot(5x^3+4x^2-3x+8)+(x^2+3x-5)^7\cdot(15x^2+8x-3)$

• So far I have: d/dx= (x^2+3x-5)^7 (15x^2+8x-3x+8)+ (5x^3+4x^2-3x+8)(14x+21x-35)^6 Is this the answer? – Kevin Sep 29 '14 at 17:39
• No, sorry, I edited my answer a bit to help you :) – rae306 Sep 29 '14 at 17:40
• The answer is the line under 'solution'. – rae306 Sep 29 '14 at 17:51
• Now I see where I messed up. 1)(2x+3) 2) (15x^2+8x-3) – Kevin Sep 29 '14 at 17:53
• $1$) requires both the power rule and the chain rule: $\frac{d}{dx}(x^2+3x-5)^7=7\cdot(x^2+3x-5)^6\cdot(2x+3)$. – rae306 Sep 29 '14 at 17:54

$f(x)= ( x^2+3x−5)^7 (5x^3+4x^2−3x+8) =u(x)v(x)$ So using $f'=uv'+u'v$ We get

$f'(x)=7( x^2+3x−5)^6(2x+3)(5x^3+4x^2−3x+8)+( x^2+3x−5)^7(15x^2+8x-3)$