Derivative of Integral with Limits of 0 and ln x

[Justabeginner]

Homework Statement

f(x)= ∫dt/(2 + sin t)
A and B are the lower and upper limits of the integral, respectively, where A is 0, and B is ln x.

Homework Equations

The Attempt at a Solution

g(x)= ln x
f(g(x))= ∫1/(2+sin t) dt, with a= 0 and b=g(x)
f'(x)= d/dx ∫1/2+sin x, with a=0, and b=x

Can't figure out what to do next, and pretty sure what I've done so far is wrong^. Any help is appreciated, thanks so much!

 

[STEMucator]

Homework Statement

f(x)= ∫dt/(2 + sin t)
A and B are the lower and upper limits of the integral, respectively, where A is 0, and B is ln x.

Homework Equations

The Attempt at a Solution

g(x)= ln x
f(g(x))= ∫1/(2+sin t) dt, with a= 0 and b=g(x)
f'(x)= d/dx ∫1/2+sin x, with a=0, and b=x

Can't figure out what to do next, and pretty sure what I've done so far is wrong^. Any help is appreciated, thanks so much!

Here's an easy way to remember it, and you can spend time proving it if you want to :

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))(b'(x)) - f(a(x))(a'(x))$$

 

[CAF123]

g(x)= ln x
f(g(x))= ∫1/(2+sin t) dt, with a= 0 and b=g(x)*
f'(x)= d/dx ∫1/2+sin x, with a=0, and b=x

The original problem was to find f'(x), right? From *, you have f(g(x)). Now apply chain rule and FTOC to find (f(g(x)))^' = d (f(g(x)))/dx.

 

[Justabeginner]

Thank you both!

Using Zondrina's equation, I get:

{[(1)/(2+sin(ln x))] * (1/x)}

Is this correct?

 

[CAF123]

Thank you both!

Using Zondrina's equation, I get:

{[(1)/(2+sin(ln x))] * (1/x)}

Is this correct?

Try what I suggested - what do you get?