# Differentiating by Partial Differentiation.

Among the methods for finding derivatives, differentiating by partial differentiation looks interesting. Is there any general proof for this method. For instance my text mentions this method. Let

$f\left(x,y\right)=x^3+y^3-3axy=c\left(constant\right)$

∴ $\frac{dy}{dx}=-\frac{f_x}{f_y}$

Where $f_x$ is the partial differentiation of the function w.r.t 'x', and $f_y$ w.r.t 'y'.

# ∴$\frac{dy}{dx}=-\frac{x^2-ay}{y^2-ax}$

Which is correct.

But in case of exponential and logarithmic functions we have a similar but different method.

Let us have $y=x^{sinx}$. Therefore according to the method of partial differentiation, we have:

$\frac{dy}{dx}=f_{sin\left(x\right)}\left(d.c\:of\:x\right)+f_x\left(d.c\:\left(sin\left(x\right)\right)\right)$

Where d.c is the differential coefficient.

So we have: $\frac{dy}{dx}=sin\left(x\right)\left(x^{sinx-1}\right)\frac{dx}{dx}+x^{sinx}logx\:\frac{d\left(sinx\right)}{dx}=sinx\left(x^{sinx-1}\right)+x^{sinx}logx\left(cosx\right)$

$\frac{dy}{dx}=x^{sinx}\left[cosx\left(logx\right)+\frac{sinx}{x}\right]$

Which is the solution we get if we do it by taking logarithm and proceeding algebraically. When I matched both up, it sort of looks like a valid method. I would like to know whether there exists any algebraic proof for this method.

• This is a consequence of the implicit function theorem. – copper.hat Sep 6 '14 at 12:55

Let $g(x,z) = x^z.$ Then $$\frac{\partial g(x,z)}{\partial x} = z x^{z - 1}.$$ $$\frac{\partial g(x,z)}{\partial z} = x^z \ln x.$$ Then the total derivative of $g(x,z)$ with respect to some parameter $t$ is $$\begin{eqnarray} \frac{dg(x,z)}{dt} &=& \frac{\partial g(x,z)}{\partial x} \frac{dx}{dt} + \frac{\partial g(x,z)}{\partial z} \frac{dz}{dt} \\ &=& z x^{z - 1} \frac{dx}{dt} + x^z \ln x \frac{dz}{dt}. \end{eqnarray}$$ Take $t = x;$ then $$\frac{dg(x,z)}{dx} = z x^{z - 1} + x^z \ln x \frac{dz}{dx}.$$ This is the same thing you get as an intermediate result, where the differential coefficient of $x$ is $\frac{dx}{dx} = 1,$ except that I've written $\frac{dz}{dx}$ where you want to write the differential coefficient of $\sin x.$
Now consider $y = x^{\sin x},$ which is simply $y = g(x,z)$ with the restriction that $z = \sin x,$ so $$\begin{eqnarray} \frac{dy}{dx} &=& \left(z x^{z - 1} + x^z \ln x \frac{dz}{dx}\right)_{z = \sin x} \\ &=& x^{\sin x - 1} \sin x + x^{\sin x} \ln x \cos x. \end{eqnarray}$$
I delayed asserting that $z = \sin x$ as long as I could, because I wanted to be able to figure out the partial derivatives first. It's hard to interpret $\frac{\partial g(x,z)}{\partial z}$ (which describes what happens if you hold $x$ constant and vary $z$) when you've already made it a fact that $z$ is a function of $x$.
Not quite the same as the implicit function theorem but using almost all the same mechanisms. Alternatively, I suppose you could define $h(x,y,z) = x^z - y,$ set $h(x,y,z) = 0,$ apply the implicit function theorem directly to get $\frac{dy}{dx}$ in terms of $x$ and $z,$ and proceed from there.