Question

Calculate the following derivatives
\[\frac{d}{dx} \int_0^x dt \cos(t^{3}+1)\]

Answer 1

\[\large \frac{d}{dx} \int_0^x dt \cos(t^{3}+1)\]

Answer 2

interesting

Answer 3

thanks i forgot to add math signs now is ok

Answer 4

use the definition from ftc

Answer 5

i need complete solution guys i have exam on friday i cant trust my own solution..

Answer 6

thanks @lgbasallote i was unable to read it .

so by using first form Fundamental theorem of calculus it is (derivative and integral gets cancelled )
cos(x^3+1)

Answer 7

use this
\[\frac{d}{dx} \int\limits_{a(x)}^{b(x)} f(t) dt= f(b(x)) \ b'(x)- f(a(x)) \ a'(x) \]

Answer 8

\[y=f'(x)\]
by using the definition of a definite integral we get
\[\int_{a(x)}^{b(x)}(y=f'(x))\to\ f(b(x))-f(a(x))\]
by using the chain rule we get
\[\frac{d}{dx}[f(b(x))-f(a(x))]=f'(b(x)b'-f'(a(x)))a'\]

Answer 9

i gots too many paranthesis lol

Answer 10

and how looks solution of my question according this formulas ?

Answer 11

well, you now f'(x) and a(x) and b(x) soo use them

Answer 12

ok thanks i take a look at it, i hope i can do ;)

Answer 13

if not, dont worry; we can then help out some more

Answer 14

ok thank you

Answer 15

\[

\frac{d}{dx} \int_0^x f(t) dt=f(x)\text{ by the FTC, so }\\
\frac{d}{dx} \int_0^x \cos(t^{3}+1) dt=\cos(x^{3}+1)

\]

Answer 16

thank you very much Mr Saab can you say me pls what FTC is ?

Answer 17

@tunahan FTC(fundamental theorem of calculus ) is it is a relation between derivative and integrals

Answer 18

@sami thank you