Implicit differentiation

In summary: = ycos(xy)+ xcos(xy)(dy/dx)= (ycos(xy)+ xcos(xy))dy/dx= (ycos(xy)+ xcos(xy))((d/dx)(sin(xy)))= (ycos(xy)+ xcos(xy))(cos(xy)(dx/dx)+ sin(xy)(dy/dx))= (ycos(xy)+ xcos(xy))(-sin(xy)(d(xy)/dx)+ cos(xy)(dy/dx))= (ycos(xy)+ xcos(xy))(-sin(xy)(y+ x(dy/dx))+ cos(xy)(dy/dx))= (ycos(xy)+ xcos(xy))(-ysin(xy)- xsin(xy)(dy/dx))+ (
  • #1
physics=world
110
0
1. y = sinxy



Homework Equations





3. this was my attempt

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




im getting stuck. i don't think I am starting it right. any suggestions.
 
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  • #2
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
jmcelve said:
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
Sorry, I'm not really following your steps. Can you maybe show me step-by-step what you're doing?
 
  • #5
use sin^-1 (y)=xy and then differentiate both sides
 
  • #6
jmcelve said:
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
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
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))
 

What is implicit differentiation?

Implicit differentiation is a method used in calculus to find the derivative of an equation where the dependent variable is not explicitly stated. It allows us to find the rate of change of a function even when it is not written in the form of y=f(x).

Why do we use implicit differentiation?

Implicit differentiation allows us to find the derivative of a function when it is not possible or convenient to solve for the dependent variable explicitly. This is particularly useful in cases where the equation is too complex to solve or when the dependent variable is not isolated.

What are the steps involved in implicit differentiation?

The first step is to differentiate both sides of the equation with respect to the independent variable. Then, we use the chain rule for any terms that involve the dependent variable. Next, we solve for the derivative of the dependent variable. Finally, we simplify the resulting equation to find the final derivative.

Can implicit differentiation be used for all types of equations?

No, implicit differentiation can only be used for equations that are implicitly defined. This means that the dependent variable is not isolated on one side of the equation. It is also important to note that implicit differentiation may not always work for equations with multiple variables or for non-differentiable functions.

What are some real-world applications of implicit differentiation?

Implicit differentiation has many applications in fields such as physics, engineering, and economics. It can be used to find the velocity and acceleration of objects in motion, to optimize production and profit in business, and to model complex systems in engineering. It is a powerful tool for analyzing and understanding relationships between variables in real-world scenarios.

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