Question

Find dy/dx.
y= cos(t^2)dt on the integral from sqrt(x) to pi/3

Answer 1

u have to dy/dx of \[\int\limits_{\sqrt{x}}^{\pi/3} y dx\]

Answer 2

where y= cos(t^2) dt

Answer 3

\[\int\limits_{g(x)}^{h(x)} f(x) dx =y\]
then dy/dx= f(g(x)) g'(x) - f(h(x)) h'(x)

Answer 4

Thank you, I'm getting confused with what to do with the cos(pi/3)^2, the cos of pi/3 is 1/2, so am I sqauring the 1/2?

Answer 5

sorry ...dy/dx= f(h(x)) h'(x) - f(g(x)) g'(x)

Answer 6

here h(x)= pi/3 so h'(x)=0

Answer 7

f(h(x))= cos[ (pi/3)^2) ]

Answer 8

so use calculator for finding cos[ (pi/3)^2) ]

Answer 9

but h'(x) =0 so f(h(x)) h'(x)=0

Answer 10

does it make sense?

Answer 11

So is my answer (cosx)/(2sqrtx)? or 1/2* cosx/sqrtx?

Answer 12

- cos(x) [(1/ {2sqrt(x)} ]

Answer 13

Oh thanks, I forgot to switch the h(x) and g(x)

Answer 14

\[\int\limits\limits_{g(x)}^{h(x)} f(x) dx =y\]
then dy/dx= f(h(x)) h'(x) - f(g(x)) g'(x)

Answer 15

Thank you for your help!

Answer 16

:)