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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.

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

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