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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?

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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

Please answer these questions first:

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

$\frac{d}{dx}(5x^3+4x^2−3x+8)=$


Solution:

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

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

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

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