By Fundamental Theorem of Calculus
G'=d/dx(∫sin^5tdt) where ∫ is from (x^2-5) to (1/x)
First you calculate:
∫sin^5dt (where ∫ is from (x^2-5) to 1/x) which will give you a function with variable x
Then you calculate the derivative with respect to x
Hunter R.
asked 06/02/21Find G'(x) by using the Fundamental Theorem of Calculus.
By Fundamental Theorem of Calculus
G'=d/dx(∫sin^5tdt) where ∫ is from (x^2-5) to (1/x)
First you calculate:
∫sin^5dt (where ∫ is from (x^2-5) to 1/x) which will give you a function with variable x
Then you calculate the derivative with respect to x
Yefim S. answered 06/02/21
Math Tutor with Experience
G'(x) = sin5(1/x)(1/x)' - sin5(x2 - 5)(x2 - 5)' = -1/x2·sin5(1/x) - 2x·sin5(x2 - 5)
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