Question

A normal line is a line perpendicular to a tangent line at the point where the tangent meets the function. Give the equation of the normal line to the graph of y=3xsqrt{x^2+6}+4 at the point (0,4)

Answer 1

so we're gonna need the calculus version of point-slope form
y-y1=f'(x1)(x-x1)
which means we need to find dy/dx (at x=x1)

then we need the fact that the normal line to this line has the slope of the negative reciprocal -1/(f'(x1)

between those facts you're good

Answer 2

so step one (as usual in calc) find dy/dx

Answer 3

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

Answer 4

or 6x/2(x^2+6)

Answer 5

or 3x/x^2+6

Answer 6

I hadn't tried yet, but that's not what I'm getting...

Answer 7

oh

Answer 8

i did the chain rule for sqrt{x^2+6}

Answer 9

oh so would it be 3x/sqrt{x^2+6}

Answer 10

Yeah, but you need to use it in conjunction with the product rul so be very careful...
I even have messed up a little according to wolfram, so I'm looking where I lost a 2

Answer 11

that's closer^

Answer 12

oh would the 3x be part of sqrt{x^2+6} in the beginning

Answer 13

would i have to use the chain rule for (3xsqrt{x^2+6}) or just sqrt{x^2+6}

Answer 14

product then chain rule
(3x)'(sqrt(x^2+6)+3x(sqrt(x^2+6))'
^^^^^^^^
chain rule here

Answer 15

hmm i kinda got this for the product rule i used the chain rule in the product rule... 3(sqrt{x^2+6})+1/2(x^2+6)^-1/2(2x)(3x)

Answer 16

yeah, that looks good
now the simplification is the annoying bit, I only just found my earlier mistake :/

Answer 17

simplify the obvious stuff
then factor out (x^2+6)^(-1/2) with finess

Answer 18

dy/dx= 3sqrt{x^2+6}+6x^2/2sqrt{x^2+6}

Answer 19

well, you could just use a calculator I suppose...
right what you have so far though, just a pain to evaluate

Answer 20

so you should keep going and simplify if you can

Answer 21

dy/dx= 3sqrt{x^2+6}+3x^2/sqrt{x^2+6}

Answer 22

you can factor out 3(x^2+6)^(-1/2) if you want to do it to look nice

Answer 23

hmm how?

Answer 24

dy/dx=3(x^2+6)^(1/2)+3x^2(x^2+6)^(-1/2)
=3(x^2+6)^(-1/2)(x^2+6+x^2)=3(2x^2+6)/sqrt(x^2+6)
=6(x^2+3)/sqrt{x^2+6}
watch the exponents carefully if that lost you

Answer 25

I have to go eat....

Evaluate that at x=0 to get the derivative at that point, which is your slope of the tangent f'(0)
the slope perpendicular to that will be the normal slope, so that is the one you want.

We know it will be -1/f'(0) from algebra.
so use -1/f'(0) in the the slope m in the point-slope formula for a line
y-y1=m(x-x1)

See you later perhaps

Answer 26

okay thank you so much! have a great dinner :) thank you.