# Implicit differentiation of a two variables function

A function $z=z(x,y)$ is given implicitly by the function $$f\left(\frac{x}{y},\frac{z}{x^{\lambda}}\right)=0$$ where $\lambda\in\mathbb{R},\lambda\neq0$.

I have to show that if $f(u,v)$ is differentiable and $$\frac{\partial f}{\partial v}(u,v)\neq0$$ then $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\lambda z$$

I tried to do this using the Implict function theorem: $$\frac{\partial z}{\partial x}=-\displaystyle\dfrac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$$ So calling $$F(x,y)=f\left(\frac{x}{y},\frac{z}{x^{\lambda}}\right)$$ and applying the chain rule, I got: $$\frac{\partial F}{\partial x}=\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial x}$$ so $$\frac{\partial F}{\partial x}=\frac{\partial f}{\partial x}\cdot\frac{1}{y}-\frac{\partial f}{\partial y}\cdot\frac{\lambda z}{x^{\lambda+1}}$$ But I don't know if is this the right way and, if it really is, what I'm supposed to do from now on?

I did all this trying to find some expression to $\frac{\partial f}{\partial x}$, but I don't know if it worked.

• Some observations: (i) You should use the fact that $F(x,y)=0$ for all $x,y$ [currently you are ignoring that fact], (ii) It is better to write "$\frac{\partial f}{\partial u}$" rather than "$\frac{\partial f}{\partial x}$" to avoid confusion and to be consistent with the notation of the question [in particular, your $\frac{\partial x}{\partial x}$ is not quite correct], (iii) you forgot that differentiating $zx^{-\lambda}$ with respect to $x$ must use the product rule and that is where you get derivatives of $z$ in the picture. Can you solve the problem now? Jul 3, 2016 at 3:32
• Also, you can get rid of the formula "$\frac{\partial z}{\partial x} = -\frac{\partial f/ \partial x}{\partial f/\partial z}$" (I'm not sure what $\partial f /\partial x$ is even intended to mean). Jul 3, 2016 at 3:35
• @Michael I fixed it, but does it equal to 0 (the last expression)? Jul 3, 2016 at 4:04
• I do not understand your comment above. What did you fix? What equals 0? If you can solve your own question now, one method is to answer your own question below. Jul 3, 2016 at 15:55
• I don't know what to do. Jul 3, 2016 at 18:39

$$df=\frac{\partial f}{\partial u}du+\frac{\partial f}{\partial v}dv=0$$ where $$du=d(x/y)=\frac{1}{y}dx-\frac{x}{y^2}dy$$ $$dv=d(z/x^\lambda)=\frac{1}{x^\lambda}dz-\frac{\lambda z}{x^{\lambda+1}}dx$$ We know that $$dz=\frac{\partial{z}}{\partial x}dx+\frac{\partial{z}}{\partial y}dy$$ So $$dv=d(z/x^\lambda)=\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}dy-\frac{\lambda z}{x^{\lambda+1}}dx$$ Now $df$ becomes $$df=\frac{\partial f}{\partial u}\bigg( \frac{1}{y}dx-\frac{x}{y^2}dy\bigg)+\frac{\partial f}{\partial v}\bigg(\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}dy-\frac{\lambda z}{x^{\lambda+1}}dx \bigg)=0$$ This equation is always satisfied when $$\frac{1}{y}dx-\frac{x}{y^2}dy=0$$ $$\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}dy-\frac{\lambda z}{x^{\lambda+1}}dx=0$$ From the first condition it follows that $$dy=\frac{y}{x}dx$$ Replacing this into the second condition $$\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}dx-\frac{\lambda z}{x^{\lambda+1}}dx=0$$ $$\bigg(\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}-\frac{\lambda z}{x^{\lambda+1}}\bigg)dx=0$$ $$\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}-\frac{\lambda z}{x^{\lambda+1}}=0$$ $$x\frac{\partial{z}}{\partial x}+y\frac{\partial{z}}{\partial y}=\lambda z$$