# Derivatives of functions of products of variables

1. Aug 3, 2013

### chipotleaway

Is there a general formula for (total) derivatives of functions of the form $f(xy(x)+z(x)$?

I tried the most simple function of that form $f(xy(x)+z(x))=xy(x)+z(x)$ and the formula I got was $\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} \frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{dz}{dx}$ though I'm unsure if it's even close to what I'm after. The formula Mathematica gave is $y\frac{\partial f}{\partial x}+x\frac{\partial f}{\partial y} \frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{dz}{dx}$

Thanks

2. Aug 3, 2013

### HallsofIvy

Staff Emeritus
You have a missing right parenthesis here. Do you mean "f(xy(x)+ z(x))"? And do you mean y(x) and z(x) as functions of x rather that y times x and z times x? Assuming that, the usual "chain rule" for functions of several variables applies:
$$\frac{df}{dx}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}+ \frac{\partial f}{\partial z}\frac{dz}{dx}$$
In this case, writing f' as the derivative of f(u) with respec to u, that becomes
$$\frac{df}{dx}= f'(xy+ z)y+ f'(xy+ z)x\frac{dy}{dx}+ f'(xy+z)\frac{dz}{dx}$$.

Well, that's what I wrote before but with f(u)= u, as here, f'= 1 so that is
$$\frac{df}{dx}= y+ x\frac{dy}{dx}+ \frac{dz}{dx}$$.

As a check you can take $$y= 2x$$ and $$z= x^2$$
Then $$f(xy+ z)= xy+ z= 2x^2+ x^2= 3x^2$$ which has derivative 6x.

In this example, dy/dx= 2 and dz/dx= 2x so the formula above would give
$$\frac{df}{dx}= y+ x\frac{dy}{dx}+ \frac{dz}{dx}= 2x+ x(2)+ 2x= 6x$$

Yes, that is precisely the formula I gave!