Derivative of Integral: Is F'(x) = 2x sin(x^2) the Correct Answer?

[theRukus]

Homework Statement

Find the derivative of the function
F(x) = \int^0_{x^2-1}\frac{sin(t+1)}{t+1}dt

Homework Equations

The Attempt at a Solution

F'(x) = -\frac{sin(x^2)}{x^2}

I'm just learning this and unsure if this is correct. It seems too easy?

 

[Mark44]

Homework Statement

Find the derivative of the function
F(x) = \int^0_{x^2-1}\frac{sin(t+1)}{t+1}dt

Homework Equations

The Attempt at a Solution

F'(x) = -\frac{sin(x^2)}{x^2}

I'm just learning this and unsure if this is correct. It seems too easy?

Right. It's not as easy as you are making it. You need to use the chain rule.

F(x) = \int^0_{x^2-1}\frac{sin(t+1)}{t+1}dt = -\int_0^{x^2-1}\frac{sin(t+1)}{t+1}dt

The Fundamental Theorem of Calculus says that, if
F(x) = \int_0^x f(t)dt
then F'(x) = f(x)

Notice however, that one of your integration limits is not x, but is instead a function of x.

\frac{d}{dx}\int_0^{u} f(t)dt = \frac{d}{du}\int_0^u f(t)dt \cdot \frac{du}{dx}

Now the integral matches the form in the FTC.

 

[theRukus]

So the answer would be,

-\frac{sin(x^2)}{x^2} \cdot 2x

Is this now correct?

 

[SammyS]

\cdot for center dot.

 

[Mark44]

So the answer would be,

-\frac{sin(x^2)}{x^2} \cdot 2x

Is this now correct?

Looks good, but can be simplified a bit.