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Find the derivative of the function. m(t)=−2t(5t3−1)8 m′(t)=

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Answer:


m'(t)=-2(5t^3-1)^8-240t^3(5t^3-1)^7

Explanation:

Assuming you mean
m(t)=-2t(5t^3-1)^8

Let the following:


f(t)=-2t\\f'(t)=-2\\g(t)=(5t^3-1)^8\\g'(t)=120t^2(5t^3-1)^7

By product rule:


m'(t)=f'(t)g(t)+f(t)g'(t)=-2(5t^3-1)^8-240t^3(5t^3-1)^7

User Heehaaw
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