Question

If $f(x) = \int_{-1}^{x} t^8 dt$ then a) $f'(x) = \frac{x^9 + 1}{9}$ b) $f'(-2) = -56.7$

Name: If f(x) = ∫-1^x t^8 dt then a) f'(x) = (x^9 + 1)/(9) b) f'(-2) = -56.7
Uploaded: 2023-06-26T00:10:21-08:00
Duration: 2 min 10 s
Channel: Adi S
Description: If f(x) = ∫-1^x t^8 dt then a) f'(x) = (x^9 + 1)/(9) b) f'(-2) = -56.7

          If $f(x) = \int_{-1}^{x} t^8 dt$ then a) $f'(x) = \frac{x^9 + 1}{9}$ b) $f'(-2) = -56.7$

Added by Michael P.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/26/2023

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We need to find the derivative of this function, f'(x), and the value of the derivative at x = -2. Show more…

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If f(x)=int_(-1)^x t^(8)dt then a) f^(')(x)= b) f^(')(-2)= 1ff= t8dt then 9+1 afx= 9 bf-2= 56.7

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If $f(x) = \int_{-1}^{x} t^8 dt$ then a) $f'(x) = \frac{x^9 + 1}{9}$ b) $f'(-2) = -56.7$

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