Fundamental theorem of calculus

[fk378]

Homework Statement

Find the derivative of cost/t (dt) evaluated from 3 to the sqrt of x

The Attempt at a Solution

using the fundamental theorem of calculus, I have [cos(sqrt x)]/(sqrtx)
I know I have to use the chain rule but I don't know how to go about it from here. Any tips?

 

[Dick]

You just have to differentiate your function of x. Start with the derivative of cos(sqrt(x)). What is it? If you can get that using the chain rule the just use the quotient rule to do the whole thing.

 

[Gib Z]

Are you trying to find the integral or the derivative? It says derivative, but "evaluated from 3 to the sqrt of x" implies integral :( Also, using FTC also implies you are finding the integral...

 

[Curious3141]

I think this is the question, basically.

Evaluate \frac{d}{dx}\int_3^{\sqrt{x}}\frac{\cos{t}}{t}dt and that's a trivial application of FTC.

Try making the substitution u= \sqrt{x}, working out \frac{d}{du}\int_c^{u}\frac{\cos{t}}{t}dt, then using the chain rule \frac{dI}{dx} = (\frac{dI}{du})(\frac{du}{dx}) (c is an arbitrary constant,the lower bound does not really matter).

 

[HallsofIvy]

Homework Statement

Find the derivative of cost/t (dt) evaluated from 3 to the sqrt of x

That makes no sense! You mean "find the derivative of the integral of cost/t dt evaluated from 3 to the sqrt(x)[/itex]

The Attempt at a Solution

using the fundamental theorem of calculus, I have [cos(sqrt x)]/(sqrtx)
I know I have to use the chain rule but I don't know how to go about it from here. Any tips?

Perhaps it would make sense if you wrote f(u)= int{3 to u} cos(t)/t dt and u(x)= sqrt(x). Now apply the chain rule to those two functions: df(x)/dx= df(u)/du du/dx. both of those should be easy now.