# Differentiation - finding equation of normal to curve

1. Oct 13, 2011

### studentxlol

1. The problem statement, all variables and given/known data

Find the equation of the normal to the curve y=(6x+3)^(1/2) at the point for which x=13.

2. Relevant equations

dy/dx=dy/du*du/dx

y-y1=m(x-x1)

3. The attempt at a solution

y=(6x+3)^(1/2)
dy/dx=1/3(6x+3)^2

at x=13, dy/dx=-3(6x13+3)^2=19683.

obviously a silly answer and the textbook disagrees.

So at this point I could simplify y=(6x+3)^(1/2) to y=(2x+1)^2 maybe? Even after doing this, I end up with an even slightly smaller silly answer and the textbook still disagrees, unless the textbook is wrong, although this is highly unlikely.

HELP!

2. Oct 13, 2011

### Hootenanny

Staff Emeritus
You will want to start by checking this:

3. Oct 13, 2011

### studentxlol

I inputted y=(6x+3)^(1/2) into wolfram alpha and the answer came out to be:

√3/(2x+1)^(1/2)

I'm confused. I used the chain rule to get:

subu=6x+3 so that y=u^1/2
dy/du=1/2u^-(1/2)=1/2(6x+3)^-(1/2) and du/dx=6

=> dy/dx=dy/du*du/dx=3(6x+3)^-(1/2).

???

Last edited: Oct 13, 2011
4. Oct 13, 2011

### LawrenceC

Work out the derivative by hand. It's straightforward.

5. Oct 13, 2011

### HallsofIvy

I put that into WolframAlpha and got the correct answer.