Differentiation - finding equation of normal to curve

[studentxlol]

Homework Statement

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

Homework Equations

dy/dx=dy/du*du/dx

y-y1=m(x-x1)

The Attempt at a Solution

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

gradient of normal = -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!

 

[Hootenanny]

You will want to start by checking this:

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

 

[studentxlol]

You will want to start by checking this:

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).

?

 

[LawrenceC]

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

 

[HallsofIvy]

I put that into WolframAlpha and got the correct answer.