Derivative of $f(x)=\int_{x}^{x^3} e^{t^2} dt$

Find $$f'(0)$$ and $$f'(\sqrt{3})$$ when $$f(x)=\int_{x}^{x^3} e^{t^2} dt$$

I know that $$\int_{}^{} e^{t^2} dt$$ doesn't return elementary functions, but I am not sure how to solve this. Can anyone give a little hint?

• – farruhota May 3 at 6:01

Recall by the Fundamental Theorem of Calculus, $$\frac{d}{dx}\int_a^xB(t)dt=B(x),\space\text{for a\in \mathbb{R}}$$

So if we have $$I=\int_{g(x)}^{h(x)}B(t)\space dt=\int_{a}^{h(x)}B(t)\space dt+\int_{g(x)}^{a}B(t)\space dt=\int_{a}^{h(x)}B(t)\space dt-\int_{a}^{g(x)}B(t)\space dt$$

We want to differentiate $$I$$, namely $$\frac{dI}{dx}=\frac{d}{dx}\left(\int_{a}^{h(x)}B(t)\space dt-\int_{a}^{g(x)}B(t)\space dt \right)=\underbrace{\frac{d}{dx}\left(\int_{a}^{h(x)}B(t)\space dt\right)}_{\frac{dI_1}{dx}}-\underbrace{\frac{d}{dx}\left(\int_{a}^{g(x)}B(t)\space dt\right)}_{\frac{dI_2}{dx}}$$

So $$\frac{dI}{dx}=\frac{dI_1}{dx}-\frac{dI_2}{dx}$$

Calculate separately:

$$\text{For \frac{dI_1}{dx}, let u=h(x) so that \frac{du}{dx}=\frac{dh}{dx}}$$

Hence $$\frac{dI_1}{dx}=\frac{d}{du}\left(\int_{h(a)}^u B(t)\space dt\right)\cdot\frac{du}{dx}=B(u)\cdot\frac{du}{dx}=B(h(x))\cdot h'(x)$$

And similarly, $$\text{For \frac{dI_2}{dx}, let u=g(x) so that \frac{du}{dx}=\frac{dg}{dx}}$$

Hence $$\frac{dI_2}{dx}=\frac{d}{du}\left(\int_{g(a)}^u B(t)\space dt\right)\cdot\frac{du}{dx}=B(u)\cdot\frac{du}{dx}=B(g(x))\cdot g'(x)$$ 

We conclude $$\frac{dI}{dx}=\frac{dI_1}{dx}-\frac{dI_2}{dx}=B(h(x))\cdot h'(x)-B(g(x))\cdot g'(x)$$ or that 

$$\frac{d}{dx}\left(f(x)\right)=\frac{d}{dx}\left(\int_{g(x)}^{h(x)}B(t)\space dt\right)=h'(x)B(h(x))-g'(x)B(g(x))$$

• $$f(x)=\int_x^{x^3}e^{t^2}dt$$
• $$h(x)=x^3$$
• $$g(x)=x$$
• $$B(t)=e^{t^2}$$
So really, you're not trying to find an antiderivative of $$y_1=e^{x^2}$$, but a derivative of $$y_2=\int_{g(x)}^{h(x)}e^{t^2}dt$$.
$$\int_{u}^{v} f'(t)\mathrm dt=f(v)-f(u)$$, we get that $$\left(\int_{u}^{v}f'(t)\mathrm dt\right)'=f'(v)v'-f'(u)u'$$. Here $$u$$ and $$v$$ are functions of $$x$$. Using this fact which is popularly called Leibniz Integral Rule, you can get the required result by setting $$f(t)=e^{t^2}$$, $$u(x)=x^2$$ and $$v(x)=x^3$$ and you'll be done in no time.