Question

Calculate the derivative:

ddivdx int_{x ^{2}}^{x ^{3}}sqrt{x}dt

Answer 1

THIS IS THE QUESTION:

\[ddivdx \int\limits_{x ^{2}}^{x ^{3}}\sqrt{x}dt\]

Answer 2

it asks to fine the derivative

Answer 3

**find**

Answer 4

i'm not able to understand the question, could you post it again

Answer 5

i get it\[d/dx(\int\limits_{x^2}^{x^3}\sqrt(t)dt)\]

Answer 6

so integrate first then take the derivative with respect to the new x values you get from the limits of integration

Answer 7

the derivative of an integration is simply the inegral argument itself.

\[f(x)=\int_{a(x)}^{b(x)}f'(t)dt=f[b(x)]-f[a(x)]\]
\[f'(x)=f[b(x)]'-f[a(x)]'=f'[b(x)]b'(x)-f'[a(x)]a'(x)\]

Answer 8

the shortcut it to cut out the middle part since you get right back to the argument in the integration to begin with

Answer 9

\[\sqrt{t}\text{ is your arguement}\]
\[\sqrt{x^3}\ 3x^2-\sqrt{x^2}\ 2x\]