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Implicit differentiation

  • #1
1. y = sinxy



Homework Equations





3. this was my attempt

d/dx = (cosxy)(sinxy(d\dx))+(xy(d/dx)




im getting stuck. i dont think im starting it right. any suggestions.
 
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Answers and Replies

  • #2
52
0
Some notes:

Remember that you're solving for dy/dx. Look for terms that will contain this.

Remember the chain rule for explicit differentiation: d/dx f(g(x)) = g'(x)*f'(g(x)). How do you apply the chain rule when you have f(g(x, y(x)))?
 
  • #3
Some notes:

Remember that you're solving for dy/dx. Look for terms that will contain this.

Remember the chain rule for explicit differentiation: d/dx f(g(x)) = g'(x)*f'(g(x)). How do you apply the chain rule when you have f(g(x, y(x)))?

is it like this?

y(d/dx) = cosxy(cosxy(d/dx))(x(d/dx))(y(1))
 
  • #4
52
0
Sorry, I'm not really following your steps. Can you maybe show me step-by-step what you're doing?
 
  • #5
7
0
use sin^-1 (y)=xy and then differentiate both sides
 
  • #6
Sorry, I'm not really following your steps. Can you maybe show me step-by-step what you're doing?
im trying to do this f(g(x, y(x)))

for the left side
 
  • #7
52
0
Your overall goal is to find dy/dx, right? So you need to apply d/dx to both sides of the equation -- d/dx (y) = d/dx (sin(xy)). The left side is simply y, so you don't need to apply the chain rule -- just take d/dx (y) = dy/dx. The right side requires the chain rule. This is where your composite function is.
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Don't write "d/dx". You mean "dy/dx" or d(xy)/dx.
When you wrote "d/dx = (cosxy)(sinxy(d\dx))+(xy(d/dx)", you meant
dy/dx= cos(xy)(d(xy)/dx)= cos(xy)((dx/dx)y+ x(dy/dx))
 

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