Question

What is the derivative of `int_(cosx)^(7x) \ cost^3 \ dt` wrt `x`?

Answer 1

`d/dx int_(cosx)^(7x) \ cost^3 \ dt = 7 cos(7x)^3 + sinx cos(cos^3x)`

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

`d/dx \ int_a^x \ f(t) \ dt = f(x)` for any constant `a`

(ie the derivative of an integral gives us the original function back).

We are asked to find:

`g'(x) = d/dx \ int_(cosx)^(7x) \ cost^3 \ dt`

Note that neither the upper or lower bounds are in the correct format for the FTOC to be applied, directly. We can manipulate the definite integral as follows:

`int_(cosx)^(7x) \ cost^3 \ dt = int_(a)^(7x) \ cost^3 \ dt - int_a^(cosx) \ cost^3 \ dt`

And so:

`g'(x) = d/dx int_(a)^(7x) \ cost^3 \ dt - d/dx int_a^(cosx) \ cost^3 \ dt` ... [A]

We have arbitrary chosen the lower limit as some arbitrary constant `a` (we could use `0` wlog, any number will do!). We can further manipulate these definite integral using a substitution and the chain rule. Let:

`u=7x => (du)/dx = 7`
`v=cosx => (dv)/dx = -sinx`

Then substituting into [A], and applying the chain rule, we get:

`g'(x) = d/dx int_(a)^u \ cost^3 \ dt - d/dx int_a^v \ cost^3 \ dt`

`" " = (du)/dx d/(du) int_(a)^u \ cost^3 \ dt - (dv)/dx d/(dv) int_a^v \ cost^3 \ dt`

`" " = 7 d/(du)int_(a)^u \ cost^3 \ dt - (-sinx) d/(dv) int_a^v \ cost^3 \ dt`

`" " = 7 d/(du)int_(a)^u \ cost^3 \ dt + sinx d/(dv) int_a^v \ cost^3 \ dt`

And now the derivative of both the integrals are in the correct form for the FTOC to be applied, giving:

`g'(x) = 7 cosu^3 + sinx cosv^3`

And restoring the initial substitution we get:

`g'(x) = 7 cos(7x)^3 + sinx cos(cos^3x)`