## Callisto 2 years ago Find $$\frac{df}{dx}$$ where $f(x) = \int_{-x}^{x^2}\frac{1}{\sqrt{1+t^6}}dt$ for x>0 How to start??

1. Callisto

brb

2. zepdrix

So this is one of those problems that involves requires you to apply the FTC Part 1. The Fundamental Theorem of Calculus part 1 is,$\huge \frac{d}{dx}\int\limits_c^x f(t) dt=f(x)$ With this problem, we have an function of x within the upper and lower limits, so it'll look a bit different.$\large \frac{d}{dx}\int\limits_{-x}^{x^2}f(t)dt=f(x^2)\color{blue}{(\frac{d}{dx}x^2)}-f(-x)\color{blue}{(\frac{d}{dx}-x)}$The blue pieces are there because we have to apply the chain rule.

3. tkhunny

Fundamental Theorem of Integral Calculus will get you started. $$f(x) = \int\limits_{-x}^{x^{2}}g(t)\;dt\;=\;G(x^{2}) - G(-x)$$ Where $$G(x)$$ is an antiderivative of $$g(x)$$. Chain Rule will get us the rest of the way. $$\dfrac{d}{dx}\left(G(x^{2}) - G(-x)\right)\;=\;g(x^{2})\cdot (2x) - g(-x)\cdot (-1)$$

4. Callisto

I'm sorry... I was so blind... Thanks all!!!!

5. tkhunny

Don't forget that we DO have to believe that the antiderivative EXISTS! Good work.

6. Callisto

f(-x) = f(x) too, I guess. Since it's an even function..