Question

Compute the derivative: d/dx integral (sqrt(t))dt from cosx to sinx

Answer 1

@sourwing

Answer 2

@ganeshie8

Answer 3

\[\dfrac{d}{dx} \int \limits_{\cos x}^{\sin x} \sqrt{t}~ dt = \sqrt{\sin x} (\sin x)' - \sqrt{\cos x} (\cos x)' = \cos x \sqrt{\sin x} + \sin x \sqrt{\cos x } \]

Answer 4

That's what I got for the answer but it's not one of the solutions. >_<

Answer 5

what are your solutions

Answer 6

Oh wait, nvm. Duh, I saw my mistake now! THanks! It was just in a different form.

Answer 7

Can you help me with another problem?

Answer 8

Let the function h(x) = integral arctan(t)dt from x to x^2. Find h ' (x)

The variables throw me off!

Answer 9

work it the same way...

Answer 10

\[\dfrac{d}{dx} \int \limits_{x}^{x^2} \arctan (t)~ dt = \arctan(x^2) (x^2)' - \arctan(x) (x)' = 2x \arctan(x^2) - \arctan(x) \]

Answer 11

Are x and t two different variables or do they represent the same thing? I always get thrown off by that.

Answer 12

the variable under definite integral is dummy, it doesnt matter whether its "t" or "p" or "q" or something else

Answer 13

\[\int \limits_a^b f(t) dt = \int \limits_a^b f(p) dp = \int \limits_a^b f(q) dq = \cdots \]

Answer 14

All integrals above evaluate to same value^

Answer 15

So, nothing to wry about what variable is there under the definite integral

Answer 16

Makes sense. :D Thanks so much for your help! You're the best.

Answer 17

\[\dfrac{d}{d\color{red}{x}} \int \limits_{\color{red}{x}}^{\color{red}{x^2}} \arctan (t)~ dt \]

Answer 18

those should match however, otherwise u wud get 0 cuz :
\(\dfrac{d}{d\color{red}{x}}f(t) = 0 \)

Answer 19

Just checked, the solutions are:
2x/1+x^2
arctan(x^2)-arctan2
2xarctan(x^2)
2x/1+x^4

Answer 20

for the previous problem ?

Answer 21

yes

Answer 22

\[\dfrac{d}{dx} \int \limits_{x}^{x^2} \arctan (t)~ dt \\ = \arctan(x^2) (x^2)' - \arctan(x) (x)' \\ = 2x \arctan(x^2) - \arctan(x)\]

Answer 23

^^thats not the answer ?

Answer 24

I couldnt find it in the solutions?

Answer 25

Not sure