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

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.

• No, the derivative is not $t^8tan(t-1)$. The rule you are talking about is called the fundamental theorem of calculus. You should do some reading here and come back if you have more questions: en.wikipedia.org/wiki/Fundamental_theorem_of_calculus – Riquelme Dec 8 '18 at 17:46
• You would have to chain rule that expression. I.e, if $F$ is the antiderivative of $t^{8}\tan(t-1)$ and $g(x) = x^2$, we have that the above integral is $F \circ g(x)$, so $(F \circ g)'(x) = F'(g(x))\cdot g'(x)$ – rubikscube09 Dec 8 '18 at 17:49

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

• Sorry but this kind of "answer" is in fact a mere "comment". An answer would at least guide the OP rather close to the solution whereas in your text, you don't even mention that the chain rule should be applied... – Jean Marie Dec 8 '18 at 17:58
• I don't have the ability to comment on a question. In any case, if you read the OP's questions, you'll notice that I actually answered all three of them. – jonan Dec 8 '18 at 18:07

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.

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