$Derivatives again. Thank you!$

Question

$Derivatives again. Thank you!$

Sorry, I still don't understand. Please help

Calculus

Hello210

221

Matchmaticians Derivatives again. Thank you!  File #1

File #1 (jpg)

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

We have

\[\frac{d}{dx}(f(x)+3x^4-4x+5)^6= 6(f(x)+3x^4-4x+5)^5 (f(x)+3x^4-4x+5)' \]
\[=6(f(x)+3x^4-4x+5)^5 (f'(x)+12x^3-4).\]
At $x=-3$
\[=6(f(-3)+3(-3)^4-4(-3)+5)^5 (f'(-3)+12(-3)^3-4)\]
\[=6(19+3(-3)^4-4(-3)+5)^5 (-6+12(-3)^3-4)\]
\[=6(19+3\cdot 81+12+5)^5 (-6-12\cdot 27-4)\]
\[=-6\times 279^5\times 334.\]
If necessary, you may compute the above number using a calculator.

Savionf

574

Hello210

0

Both were wrong. I get one more try
- Savionf
  
  0
  
  A minus sign was missing in the final answer. Please try again. It should work now.
Hello210

0

It didn’t work. The answer was 6(19+260)^5(-6-328). I ran out of fund, I’ll try to redo a different one on my own
- Savionf
  
  0
  
  My answer is the same as 6(19+260)^5(-6-328). Probably it is did not accept it because of the format of the answer. 6(19+260)^5(-6-328)=6(279)^5(-334), same as my answer.

$Derivatives again. Thank you!$

Answer

Related Questions

Search