Question

http://www.webassign.net/cgi-bin/symimage.cgi?expr=y%3Dint_%281%29%5E%28cos%5C%28x%5C%29%29%20%285%2Bv%5E4%29%2A%2A%286%29dv

Answer 1

i need to fin y'....i thought it would be (5+v^4)^6

Answer 2

and also this as well.

http://www.webassign.net/cgi-bin/symimage.cgi?expr=h%28x%29%3Dint_4%5Ex%2A%2A2%20sqrt%285%2Br%5E3%29%20dr

Answer 3

find h'(x)

Answer 4

Well you know that if you have:
\[\frac{d}{dx}\int\limits_{a}^{u(x)}f(x)dx=f(u(x))*u'(x)\]
So f(x)=(5+v^4)^6
So f(u(x))=(5+cos^4(x))^6
and u'(x)=-sin(x)
so your answer should read:
\[-\sin(x)(5+\cos^4(x))^6\]

Answer 5

For the second one you have:
f(x)=sqrt(5+r^3)
f(u(x))=sqrt(5+(x^2)^3)
u'(x)=2x
So:
\[h'(x)=(2x)\sqrt{5+x^6}\]
Using the same formula as above. (The fundamental theorem of calculus)

Answer 6

Does that make sense?

Answer 7

i will keep that in record thanks.. yes now im getting the idea..iwas just not working at first

Answer 8

Okay. The fundamental theorem of calculus is a powerful tool. There is a technique of integration called "differentiation under the integral sign" and you can move the differential operator (d/dx) in and out of the integral to make them easier to solve. Its a fairly advanced method xDD

Answer 9

ok cool thanks agian