# Find the derivative of the integral from 0 to x^4 (sec(t)dt)

1. Oct 11, 2006

### sapiental

find the derivative of the integral from 0 to x^4 (sec(t)dt)

I get sec(x^4)(4x^3)

I understand the procedure for solving this integral. However, I don't understand why exactly we need to use the chain rule in this situation.

Why isn't the answer just sec(x^4)?

Thank you!

2. Oct 11, 2006

$$\frac{d}{dt} \int_{0}^{x^{4}} \sec t dt$$. Basically you are getting a derivative. If its a derivative then you have to use the chain rule.

Last edited: Oct 11, 2006
3. Oct 11, 2006

### sapiental

d/dx integral 0 to ln(x) (sint + t^3/2)dt

using the chain rule and ftc1 I get:

u = ln(x) du/dx = 1/x

d/du (sinu + u^3/2)du/dx

sin(lnx)+(lnx)^3/2 (1/x)

is this correct? Thanks again.

4. Oct 11, 2006

$$\frac{d}{dx} \int_{0}^{\ln x} (\sin t + t^{\frac{3}{2}}) dt$$. So the derivative so far, is $$\sin(\ln x) + (\ln x)^\frac{3}{2} dt$$. So then it would be:

$$\frac{\sin(\ln x)}{x} + \frac{(\ln x)^{\frac{3}{2}}}{x}$$.

Last edited: Oct 11, 2006
5. Oct 11, 2006

### sapiental

so my answer was off by not multiplying both parts by 1/x? Thanks.

Last edited: Oct 11, 2006
6. Oct 11, 2006

7. Oct 11, 2006

### sapiental

ok ok, thanks!!

8. Oct 11, 2006

yeah, both parts need to be multiplied by $$\frac{1}{x}$$. They both contain u and thus must contain du .