- #1

- 16

- 0

**1. Homework Statement**

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

**2. Homework Equations**

Chain Rule

[tex]\frac{dy}{dx}[/tex] = [tex]\frac{dy}{du}[/tex] x [tex]\frac{du}{dx}[/tex]

**3. The Attempt at a Solution**

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

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

[tex]\frac{dy}{dx}[/tex] = 5u(5u^2 - 3)^-1/2 x -6x(3x+1)^-2

= -30xu(5u^2 - 3)^-1/2 x (3x+1)^-2

= [tex]\frac{-60x^2}{3x+1}[/tex](5([tex]\frac{2x}{3x+1}[/tex])^2 - 3)^-1/2 x (3x+1)^-2

Is that the final simplified answer?