Please Help! Let f(x)=(13x^3−3x+8)((3x^−2)+(2x^−3)). Evaluate f′(x) at the following points: f'(-6)
What is the derivative of f?
(I would use the product rule, rather than expanding, but that is up to you)
Tell me: \(1)\) The derivative of: \(\color{black}{ \displaystyle 13x^3−3x+8 }\) is? \(2)\) The derivative of: \(\color{black}{ \displaystyle 3x^{-2}+2x^{-3} }\) is?
I did the product rule and the plugged in the -6 and my answer was 39.6049
can you tell me what you got for the derivative of the function?
((39x^2)-3)((3x^-2)+(2x^-3))+((13x^3)-3x+8)((-6x^-3)+(-6x^-4))
yes, that is correct
Now we plug in x=-6... we can use wolfram for this:)
okay so is my answer the same?
because whenever I type the answer into the program I am using for my homework its says it's incorrect
39.3796
is what wolfram gives
I honestly have no clue about what happened. We got the correct derivative and I used the calculator for f'(-6)....
I know .-. I've attempted this problem 15 times
Yes, that is pretty strange. to me at least.
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