Find $\frac{df}{dx}$ where $$f(x) = \int_{-x}^{x^2}\frac{1}{\sqrt{1+t^6}}dt$$ for x0 How to start??

Question

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

callisto · Answer

brb

zepdrix · Answer

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.

tkhunny · Answer

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.

$\dfrac{d}{dx}\left(G(x^{2}) - G(-x)\right)\;=\;g(x^{2})\cdot (2x) - g(-x)\cdot (-1)$

callisto · Answer

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

tkhunny · Answer

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

callisto · Answer

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