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.