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?



Use the product rule:


Can you take it from here?


Please answer these questions first:





  • $\begingroup$ 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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.