# Taking the derivative of an integral with x in the function and in the bound

I know how you would do this: $${d\over dx}\int_2^{x^4} \tan (t^2)\,dt,$$ but how would you do this: $${d\over dx}\int_2^{x^4} \tan (x^2)\,dx.$$ I am confused on the argument $x$ being in the integral and as the bound of the integral.

Edit: Since everyone is solving in terms of $t$ as the inside, I will assume that it is in terms of $t$, not $x$. I was confused because I was unsure if there were another method to solve the second integral, or if it was just a misprint. I will assume that this is a misprint and carry on. Thanks!

• The second expression is not well-written; one should not use $x$ to represent both a dummy variable as well as something separate. The most reasonable interpretation of the second expression is as the same as the first expression, but again the second expression is ambiguous/ill-defined and should not be written that way. – angryavian Sep 15 '18 at 22:44
• It would seem that the text the second integral was obtained from has made an error in printing. – Leucippus Sep 15 '18 at 22:48

Using Leibniz Formula $$\frac{d}{dx} \left (\int_{a(x)}^{b(x)}f(x,t)\,dt \right) = f\big(x,b(x)\big)\cdot \frac{d}{dx} b(x) - f\big(x,a(x)\big)\cdot \frac{d}{dx} a(x) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x} f(x,t) \,dt.$$ where $$a(x) = 2$$ and $$b(x) = x^4$$ and $$f(x,t) = \tan t^2$$ You get $$f(x,b(x)) = \tan x^8$$ $$\frac{d}{dx}b(x) = 4x^3$$ Also $$f(x,a(x)) = \tan 2^2 = \tan 4$$ but $$\frac{d}{dx}a(x) = 0$$ and also notice that $$\frac{\partial}{\partial x} f(x,t) = \frac{\partial}{\partial x} \tan t^2 = 0$$ So, we get that $${d\over dx}\int_{2}^{x^4} \tan (t^2)dt, = 4x^3 \tan x^8$$

Note that $t$ is a dummy variable so both integrals are the same.

$$\int_{2}^{x^4} \tan (x^2)dx= \int_{2}^{x^4} \tan (t^2)dt$$

This is a composite function situation. The upper limit of integration is the inner function.

The derivative is found by chain rule, which is derivative of the outer function multiplied by derivative of the inner function.

Therefore you answer is $${d\over dx}\int_{2}^{x^4} \tan (t^2)dt= (4x^3)\tan (x^8)$$

$${d\over dx}\int_{2}^{x^4} \tan (t^2)dt,$$

Let $$F(u)= \int_2^u \tan (t^2)dt$$

Then you want to compute $F'(x^4)$.

\begin{align} F'(x^4) &= \left. {dF \over du} \cdot {du \over dx}\right|_{u=x^4} \\ &=\tan(x^8) \cdot 4x^3 \\ &= 4x^3 \tan(x^8) \end{align}

$$I=\int_2^{x^4}\tan(t^2)dt$$ and we want to find $\frac{dI}{dx}$ $$\frac{d}{dx}\int_2^{x^4}\tan(t^2)dt=\frac{d}{dx}\left[x^4\right]*\tan\left((x^4)^2\right)=4x^3\tan(x^8)$$ and we know this from the full version of the Leibniz integral rule: $$\frac{d}{dx} \left (\int_{a(x)}^{b(x)}f(x,t)\,dt \right) = f\big(x,b(x)\big)\cdot \frac{d}{dx} b(x) - f\big(x,a(x)\big)\cdot \frac{d}{dx} a(x) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x} f(x,t) \,dt$$