How to find the derivative using the chain rule?

[chaosblack]

Homework Statement

Find \frac{dy}{dx} if y = \sqrt{5u^2 -3} and u = \frac{2x}{3x+1}

Homework Equations

Chain Rule
\frac{dy}{dx} = \frac{dy}{du} x \frac{du}{dx}

The Attempt at a Solution

\frac{dy}{du} = 5u(5u^2 - 3)^-1/2

\frac{du}{dx} = -6x(3x+1)^-2

\frac{dy}{dx} = 5u(5u^2 - 3)^-1/2 x -6x(3x+1)^-2
= -30xu(5u^2 - 3)^-1/2 x (3x+1)^-2
= \frac{-60x^2}{3x+1}(5(\frac{2x}{3x+1})^2 - 3)^-1/2 x (3x+1)^-2

Is that the final simplified answer?

 

[G01]

Check your work for du/dx. I think you may have an error. You should have a 2 in the numerator, not a -6x.

 

[chaosblack]

u = \frac{2x}{3x+1}
u = (2x)(3x+1)^-1
u' = -2x(3x+1)^-2 x (3)
u' = -6x(3x+1)^-2

Did I do something wrong?

 

[CompuChip]

\frac{d}{dx}(f(x) g(x)) = \frac{df(x)}{dx} g(x) + f(x) \frac{dg(x)}{dx}
You have a term missing.

 

[chaosblack]

u = (2x)(3x+1)^-1
u' = 2(3x+1)^-1 + (2x)(-1)(3x+1)^2(3)
= 2(3x+1)^-1 - 6x(3x+1)^-2

This?

 

[G01]

Yes, that's the correct du/dx.

 

[chaosblack]

dy/dx = 5u(5u^2 - 3)^-1/2 x (2(3x+1)^-1 - 6x(3x+1)^-2)

From here do I just sub in for u?

 

[G01]

Yep. You got it.