How do I find v' for functions v=x+y using implicit differentiation?

[daster]

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

 

[dextercioby]

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

Daniel.

PS.Tell me what u get.

 

[HallsofIvy]

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.

 

[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?

 

[Galileo]

It comes from the chain rule again:

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

 

[daster]

Dank je.

 

[Galileo]

Graag gedaan.