OpenStudy (mathmusician):

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

1 year ago
OpenStudy (mathmusician):

1 year ago
OpenStudy (mathmusician):

There is an easier way to see the question.

1 year ago
OpenStudy (therawbugeyes):

do u still need help

1 year ago
OpenStudy (mathmusician):

Yes

1 year ago
OpenStudy (mathmusician):

@therawbugeyes

1 year ago
OpenStudy (s4sensitiveandshy):

Implicit differentiation

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (mathmusician):

OKay

1 year ago
OpenStudy (mathmusician):

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

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (mathmusician):

What happened to the y

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (mathmusician):

Could you write that out for me please?

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (mathmusician):

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

1 year ago
OpenStudy (s4sensitiveandshy):

hmm well what would you divide ?

1 year ago
OpenStudy (mathmusician):

The dy/dx

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (mathmusician):

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

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (mathmusician):

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

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (s4sensitiveandshy):

don't forget simple math :P

1 year ago
OpenStudy (mathmusician):

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

1 year ago
OpenStudy (s4sensitiveandshy):

looks good

1 year ago
OpenStudy (s4sensitiveandshy):

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

1 year ago
OpenStudy (mathmusician):

Thanks for the help

1 year ago
OpenStudy (s4sensitiveandshy):

np

1 year ago
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