# Implicit differentiation

1. Jan 15, 2005

### daster

Say we have two functions of x, v and y, such that v=x+y. How can I find v'?

2. Jan 15, 2005

### dextercioby

Why doncha differentiate its definition wrt to "x"??

Daniel.

PS.Tell me what u get.

3. Jan 15, 2005

### HallsofIvy

Staff Emeritus
Since v and y are functions of x, I presume by " v' " you mean "the derivative of v with respect to x", i.e. dv/dx.

Use the chain rule: v= x+ y so dv/dx= dx/dx+ dy/dx= 1+ dy/dx. What dy/dx is depends, of course, on what function y is of x.

4. Jan 16, 2005

### daster

Oh, so it's only dy/dx? Cause I remember my book doing something like (dy/dx)(dv/dx) or something. Thanks HallsofIvy.

Another question. How do I find d(e^u dy/dx)/du, where u and y are functions of x?
My book says:

$$e^u \frac{dy}{dx} + e^u \frac{d^2y}{dx^2} \cdot \frac{dx}{du}$$

I understand this is the product rule, but where'd the dx/du come from?

Last edited by a moderator: Jan 16, 2005
5. Jan 16, 2005

### Galileo

It comes from the chain rule again:

$$\frac{d}{du} \frac{dy}{dx}=\frac{d^2y}{dx^2}\frac{dx}{du}$$

6. Jan 16, 2005

### daster

Dank je.

7. Jan 17, 2005

### Galileo

Graag gedaan.