In what situation is the derivative of the integral not simply the function inside the integral?

Question

For example: $$\int_{1}^{x^2}t^8\tan(t-1)dt$$

Is the derivative of the above simply: $t^8\tan(t-1)$?

And in general, in what cases can you not apply this rule? Does it depend on the lower and upper bounds on the integral at all (doesn't seem like it should as this rule applies to indefinite integrals as well).

Also, I call it a "rule", but I realize it's just a consequence of the definitions of integrals and derivatives. In any case, in what situations should I think twice about applying this?

Any help is appreciated.

Answer 1

No, the derivative above is not what you have written there. If your boundary was simply 1 to x, this would be the case.

Read up on the Fundamental theorem of calculus in 1-D. To answer your question, you should be careful about this whenever your bounds contain something other than constants and the variable with respect to which you are deriving.

Answer 2

If you set

$$f(x)=\int_{1}^{x^2}t^8\tan(t-1)dt=\int_{1}^{x^2}g(t)dt$$

you can only get the derivative below

$$\dfrac{df(x)}{dx}=g(x^2)\dfrac{dx^2}{dx}=2xg(x^2)=2x^{17}\tan(x^2-1)$$

same as the chain rule of derivative in combined functions.

Answer 3

$$t^8\tan(t-1)=\frac{\mathrm d}{\mathrm dt}\int_{1}^{t}t^8\tan(t-1)\mathrm dt$$ Generally, $$F(x)=\int_{g(x)}^{h(x)}f(t)\mathrm dt$$ $$F'(x)=f(h(x))h'(x)-f(g(x))g'(x)$$