Calculus Help Please!

Printable View

January 27th 2009, 01:43 AM
juicysharpie

Calculus Help Please!

I'm studying for a test, and I don't understand how to do this problem. Could anyone help me?

Find the derivative dy/dx if y equals the integral from sin(x) to 5 of 1/u^3 du. Be sure to show your work and give the statement of any theorems that you use.
January 27th 2009, 03:55 AM
james_bond

I can tell you the answer: $-\cot x \csc^2 x$ .
January 27th 2009, 03:57 AM
mr fantastic

Quote:

Originally Posted by juicysharpie

I'm studying for a test, and I don't understand how to do this problem. Could anyone help me?

Find the derivative dy/dx if y equals the integral from sin(x) to 5 of 1/u^3 du. Be sure to show your work and give the statement of any theorems that you use.

$y = \int_{\sin x}^5 \frac{1}{u^3} \, du = - \int^{\sin x}_5 \frac{1}{u^3} \, du$ .

Let $t = \sin x$ and use the chain rule:

$\frac{dy}{dx} = \frac{d}{dt} \left[- \int^{t}_5 \frac{1}{u^3} \, du \right] \cdot \frac{dt}{dx}$

$= -\frac{1}{t^3} \cdot \cos x = -\frac{1}{\sin^3 x} \cdot \cos x = \frac{\cos x}{\sin^3 x}$ .

Your job is to identify and name the theorems that I used.
January 27th 2009, 04:56 AM
Rincewind

Quote:

Originally Posted by mr fantastic

$= -\frac{1}{t^3} \cdot \cos x = -\frac{1}{\sin^3 x} \cdot \cos x = \frac{\cos x}{\sin^3 x}$ .

Looks good except you dropped a minus sign on the last step.
January 27th 2009, 11:04 AM
juicysharpie

Quote:

Originally Posted by mr fantastic

$\frac{dy}{dx} = \frac{d}{dt} \left[- \int^{t}_5 \frac{1}{u^3} \, du \right] \cdot \frac{dt}{dx}$

I'm sorry, could you please explain how you got to this step? I don't quite understand how you did it.

Thanks for helping!!
January 27th 2009, 11:15 AM
mr fantastic

Quote:

Originally Posted by juicysharpie

I'm sorry, could you please explain how you got to this step? I don't quite understand how you did it.

Thanks for helping!!

Review chain rule in your class notes or textbook.