Derivative of integral?

Question

When asked questions of the type;

What is the derivative of $f(x) = \int_0^{x^2} \frac{cos(t)}{t+1}dt $

... what is the general method to solve them? Above is just an example from my workbook. I know what the answer is, I just don't know why it is what it is.

Ty. Not quite sure I understand what's going on here.

Use the fundamental theorem of calculus along with the chain rule. You need the chain rule because of the $x^2$ in the upper limit of the integral. — Paul Sundheim, Oct 20 '14 at 22:04
If you want a general method for "writing down" the answer, see the comments following this answer. — Dilip Sarwate, Oct 20 '14 at 22:07

score 1 · Answer 1 · answered Oct 20 '14 at 22:16

1

Fundamental theorem of calculus say that

$$ \frac{d }{dx} \int\limits_a^x f(s) d s = f(x)$$

Now, using chain Rule,

$$ \frac{d }{dx} \int\limits_a^{g(x)} f(s) d s = f(g(x))\cdot g'(x)$$

answered Oct 20 '14 at 22:16

score 0 · Answer 2 · answered Oct 20 '14 at 22:30

maybe an informal argument can help to gain better intuitive understanding: $$ \frac{ f(x+dx)-f(x)}{dx} = (dx)^{-1}\int_{x^2}^{(x+dx)^2} \frac{\cos(t)}{t+1}dt \\ $$ since the integrand is differentiable $\exists \theta \in [x^2,(x+dx)^2]$ such that the RHS is: $$ (dx)^{-1}((x+dx)^2-x^2)) \frac{\cos(\theta)}{\theta+1} \\ =(2x+dx)\frac{\cos(\theta)}{\theta+1} \\ \to 2x \frac{\cos x^2}{x^2+1} $$

Derivative of integral?

2 Answers2