Answer:
y' = -4eˣ sin⁵(eˣ)
Explanation:
Part 1 of the Fundamental Theorem of Calculus simply says that the definite integral of a function is equal to the antiderivative evaluated between the limits.
∫ₐᵇ f(x) dx = F(b) − F(a), where F(x) = ∫ f(x) dx.
y = ∫ₑₓ⁰ 4 sin⁵(t) dt
To integrate, use Pythagorean identity.
y = 4 ∫ₑₓ⁰ (1 − cos²(t))² sin(t) dt
Ignoring the limits for the moment, let's say u = cos(t) and du = -sin(t) dt.
y = 4 ∫ (1 − u²)² (-du)
y = -4 ∫ (1 − 2u² + u⁴) du
y = -4 (u − ⅔u³ + ⅕u⁵) + C
y = -4 (cos(t) − ⅔ cos³(t) + ⅕ cos⁵(t)) + C
Evaluate between t = eˣ and t = 0.
y = -4 (cos(0) − ⅔ cos³(0) + ⅕ cos⁵(0)) + C − [-4 (cos(eˣ) − ⅔ cos³(eˣ) + ⅕ cos⁵(eˣ)) + C]
y = -4 (1 − ⅔ + ⅕) + 4 (cos(eˣ) − ⅔ cos³(eˣ) + ⅕ cos⁵(eˣ))
Now take derivative with respect to x.
y' = 0 + 4 (-sin(eˣ) eˣ − 2 cos²(eˣ) (-sin(eˣ) eˣ) + cos⁴(eˣ) (-sin(eˣ) eˣ))
y' = -4eˣ sin(eˣ) (1 − 2 cos²(eˣ) + cos⁴(eˣ))
y' = -4eˣ sin(eˣ) (1 − cos²(eˣ))²
y' = -4eˣ sin⁵(eˣ)
This can be more easily calculated using the Second Fundamental Theorem of Calculus.