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.