How do you do this question?

y' = -4eˣ sin⁵(eˣ)

Step-by-step 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.

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents