Question

need help on the attachment @Calculus1

Answer 1

is that
\[\sqrt{x}\] as the upper limit?

Answer 2

yes

Answer 3

can't really see it, but if so, replace t in your integrand by
\[\sqrt{x}\] everywhere you see it, and then multiply by
\[\frac{1}{2\sqrt{x}}\]

Answer 4

this is like a composite function
\[F(g(x))\] where
\[g(x)=\sqrt{x}\] and
\[F(x)=\int_4^x\frac{6\cos(t)}{t}dt\]
so that
\[F(g(x))=\int_4^{x^2}\frac{6\cos(t)}{t}dt\]

Answer 5

then you get your derivative by the chain rule. so as i said replace each "t" in the integrand by
\[\sqrt{x}\] and then multiply by the derivative of
\[\sqrt{x}\] which is
\[\frac{1}{2\sqrt{x}}\]

Answer 6

where did x^2 come from? did you use u subsitution

Answer 7

ooooooooooooooh sorry

Answer 8

that was a typo my mistake

Answer 9

where did the square root go, did you use u substitution?

Answer 10

i meant to write
\[F(g(x))=\int_4^{\sqrt{x}}\frac{6\cos(t)}{t}dt\]

Answer 11

my mistake.

Answer 12

what happen to the 4

Answer 13

you mean when you take the derivative?

Answer 14

the 4 is unimportant in taking the derivative. the lower limit of integration (assuming it is a number) just determines the constant, and the derivative of a constant is 0

Answer 15

is the anti-derivative or derivative

Answer 16

ok lets go slow, because i think this might be unclear

Answer 17

ok

Answer 18

if i put
\[F(x)=\int_a^xf(t)dt\] i have a function of x. that much is clear yes?

Answer 19

yes

Answer 20

for a simple example i could write
\[F(x)=\int_2^x 2t dt\]

Answer 21

then
\[F(5)=\int_2^52tdt\] which is a number, easy to evaluate

Answer 22

but
\[F(x)=\int_2^xf(t)dt\] is not a number, it is a function of x

Answer 23

and the derivative of the integral is the integrand, so i know
\[F'(x)=2x\]

Answer 24

all i did was replace t in the integrand by x, and that is all. notice that i am not at all concerned about the lower limit of integration. it is a number and has nothing to do with taking the derivative

Answer 25

another example would be
\[F(x)=\int_0^x\cos(t)dt\] and here
\[F'(x)=\cos(x)\]

Answer 26

this is the engine behind the fundamental theorem of calculus. in fact this is the fundamental theorem of calculus in one version. so a problem like "find the derivative of this integral" is very easy. just write the integrand with the "dummy variable" replaced by x

Answer 27

the only complication comes if the limit of integration is not x, but rather some function of x, and in that case you have to use the chain rule because you are working with a composite function

Answer 28

so if
\[F(x)=\int_0^x\cos(t)dt\] then
\[F'(x)=\cos(x)\] but if
\[F(x)=\int_0^{x^4}\cos(t)dt\] then
\[F'(x)=\cos(x^4)\times 4x^3\] by the chain rule

Answer 29

"derivative of the outside function (the integral) evaluated at the inside function, times the derivative of the inside function"

Answer 30

ok so the answer to the problem I posted is 3sin*sqrt(x)/x

Answer 31

let me look again

Answer 32

no do not take the derivative of the "integrand"

Answer 33

the derivative of the integral IS the integrand

Answer 34

so it is
\[\frac{6\cos(\sqrt{x})}{\sqrt{x}}\times \frac{1}{2\sqrt{x}}\]

Answer 35

which of course cleans up to
\[\frac{3\cos(\sqrt{x})}{x}\]

Answer 36

oh so you only take the derivative of squrt, and just plug the highest limit into the t variable

Answer 37

that is right

Answer 38

Thanks satellite your still my favorite on here

Answer 39

assuming of course that the upper limit is a function of x

Answer 40

yw, and glad to help!