Question

The slope of the line normal to the curve e^x-x^3+y^2=10 at the point (0,3) is:

Answer 1

There is an easier way to see the question.

Answer 2

do u still need help

Answer 3

Yes

Answer 4

@therawbugeyes

Answer 5

Implicit differentiation

Answer 6

take the derivative of the given equation with respect to x. And then substitute (x,y) with (0,3)

Answer 7

OKay

Answer 8

Would the derivative be \[e ^{x}-3x ^{2}\]

Answer 9

looks good so far
with respect to x so \[e^x -3x^2 \frac{dx}{dx}\]
dx cancels out so left with e^x-3x^2

Answer 10

What happened to the y

Answer 11

take the derivative of y. the term y^2 is the part of left side expression.

Answer 12

Could you write that out for me please?

Answer 13

\[e^x -3x^2 +2y \frac{dy}{dx}=0\]
derivative of constant =0
now solve for dy/dx

Answer 14

ummm do i just move it to the other side?(Divide)

Answer 15

hmm well what would you divide ?

Answer 16

The dy/dx

Answer 17

if you divide first by dy/dx
you have to divide all terms at left side by dy/dx
\[\frac{e^x}{dy/dx} -\frac{3x^2}{dy/dx} +2y =0\]
like this.
looks messssy

Answer 18

first move (e^x-3x^2) to the right side. move all the terms except dy/dx from left side.

Answer 19

okay so it would look \[\frac{ dy }{ dx } = -(e ^{x}-3x ^{2})\]like

Answer 20

hmm what about 2y ?? what did you do with that ?

Answer 21

it should be \[2y\frac{ dy }{ dx } = -(e ^{x}-3x ^{2})\]
divide both sides.

Answer 22

\[\frac{ dy }{ dx }=-(e ^{x}-3x ^{2}+2y)\]

Answer 23

nahh i said divide
2y times dy/dx
^mutiplication
do the opposite of multiplication to cancel out 2y.
Don't

Answer 24

don't forget simple math :P

Answer 25

-(e^x-3x^2)/2y

Answer 26

looks good

Answer 27

Substitute x for 0 and y for 3
that's it :)

Answer 28

Thanks for the help

Answer 29

np