hope this formats :)

$$\cfrac{d}{dx}\int_{0}^{x^4}cos(\sqrt{t})dt$$

Question

hope this formats :)

$$\cfrac{d}{dx}\int_{0}^{x^4}cos(\sqrt{t})dt$$

amistre64 · Answer

$$\cfrac{d}{dx}\int_{x^4}^{0}cos(\sqrt{t})dt$$

zarkon · Answer

$$\frac{d}{dx}\int\limits_{x^4}^{0}\cos(\sqrt{t})dt=-\frac{d}{dx}\int\limits_{0}^{x^4}\cos(\sqrt{t})dt=-\cos(\sqrt{x^4})4x^3$$

anonymous · Answer

yes flip and take derivative using chain rule and FTC

anonymous · Answer

get exactly what Zarkon wrote

anonymous · Answer

although you can probably do something with 
$$\sqrt{x^4}$$

amistre64 · Answer

there zarkon is ... had to f5 it to get to see it :)

I kept thining it was:

$$2x\ cos(x^2)$$

anonymous · Answer

AMISTRE! :)

amistre64 · Answer

howdy skittl

amistre64 · Answer

$$\frac{d}{dx}\int_{a}^{b}F'(x)dx=F'(b)b'-F'(a)a'$$

i just thought since it was cos(x^2) that a' was 2x :)

anonymous · Answer

no the "inside" function is x^4

anonymous · Answer

it is not 
$$\frac{d}{dx}\int _a^b$$ it is 
$$\frac{d}{dx}\int_a^{g(x)}$$

anonymous · Answer

you are using ftc part in other form. the one you need says "derivative of integral is integrand"

anonymous · Answer

good morning!

amistre64 · Answer

right, I had some wires crossed on the test I took in class :)  i marked the right answer simply becasue it was tho only one that was close to my fauxpaux