Can someone please help explain this concept and how to solve questions like this?

Question

If f(x) = ∫ [from 4 to x⁴+8x³+6] sin(t)dt then, f'(x)= ?

Thanks in advance!

Answer 1

Your function f(x) = ∫ (from t=4 to t=x⁴ + 8x³ + 6) sin(t) dt

can be written as the composition of two functions:

f(x) = g(h(x)), or h = g ° h, where

h(x) = x⁴ + 8x³ + 6

g(x) = ∫ (from t=4 to t=x) sin(t) dt

The derivatives of the two little functions h and g are easy:

h'(x) = 4x³ + 24x²

g'(x) = sin(x) (this is just the Fundamental Theorem of Calculus)

Now we can use the Chain Rule to get the derivative of f:

f'(x) = g'(h(x))h'(x)

= sin(h(x))(4x³ + 24x²)

= (sin(x⁴ + 8x³ + 6))(4x³ + 24x²)

Thanks and good luck.

Answer 2

If F(x) = ∫_{(a to g(x))} f(t)dt, then, by the Chain Rule and Fundamental Theorem of Calculus,

F'(x) = f(g(x))g'(x).

So, if F(x) = ∫_{(4 to x}4_+8x3₊₆₎ sintdt, then F'(x) = sin(x⁴+8x³+6)(4x³+24x²)