Implicit Differentiation: How to Find the Derivative of an Unknown Function?

[cd246]

Homework Statement

I just got started on this, and am not grasping the WHOLE idea.
1.xy=25 The answer says -y/x
2.x^2+3xy+y^2=15 And this says -y^2/x^2

Homework Equations

1. dy/dx(xy)= dy/dx(25)
1=0 ?
2.dy/dx x^2+3xy+y^2= dy/dx 15
2x+3+y(dy/dx) =0

The Attempt at a Solution

2. -2x-3/y (which i know is not right.)

 

[nrqed]

Homework Statement

I just got started on this, and am not grasping the WHOLE idea.
1.xy=25 The answer says -y/x
2.x^2+3xy+y^2=15 And this says -y^2/x^2

Homework Equations

1. dy/dx(xy)= dy/dx(25)
1=0 ?
2.dy/dx x^2+3xy+y^2= dy/dx 15
2x+3+y(dy/dx) =0

The Attempt at a Solution

2. -2x-3/y (which i know is not right.)

The basic idea is this: Consider y as a function of x.

For example, you have x times y(x) = 25.

Now differentiate both sides with respect to x. You may use the chain rule on the product x times y(x).

$$\frac{d}{dx} ( x y(x)) = \frac{d}{dx} (25)$$

Apply the chain rule, take the derivatives and then isolate dy/dx.

 

[cd246]

I don't know if I got this right.
I used the product rule: (x)(y(x))'+(x)'(y(x)) and for derivative of y(x), I used the
chain rule to find the derivative. And got(x)(1*x*x)+(x)(y(x)). -> x^3+xy(x)d/dx=0 What else am I doing wrong?

 

[nrqed]

I don't know if I got this right.
I used the product rule: (x)(y(x))'+(x)'(y(x)) and for derivative of y(x), I used the
chain rule to find the derivative. And got(x)(1*x*x)+(x)(y(x)). -> x^3+xy(x)d/dx=0 What else am I doing wrong?

You can't use the chain rule to find the derivative of y(x) since it's an unknown function. Just leave it as $y'$! Now, using (x)' = 1, you get

y + x y' = 0 so y' = -y/x

It's that simple!