# differentiability of argument function

Suppose I have two functions $$f:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$$, $$g:\mathbb{R}\rightarrow\mathbb{R}$$ which are differentiable and I define $$\pi:\mathbb{R}\rightarrow\mathbb{R}$$ as $$f(x-\pi(y), y)=g(x).$$ I have to show that $$\pi$$ is differentiable wrt. $$y$$ and find the derivative of $$\pi$$ in terms of $$f$$ and $$g$$.

I believe that $$\pi$$ is differentiable by Chain rule and bc of composition of differentiable functions is differentiable. But I fail to compute the derivative of $$\pi$$ using Chain rule.

Can anybody please help me. Thank you.

• If $f=1$ and $g=1$, $\pi$ can be anything. Hence, this is not a definition and $\pi$ need not be differentiable. – amsmath Oct 21 '19 at 21:46

## 1 Answer

Define $$\phi : \mathbb R^2\to\mathbb R^2$$ by $$\phi(x,y) = (x-\pi(y),y)$$. Then $$\phi'(x,y) = \begin{pmatrix}1&-\pi'(y)\\0&1\end{pmatrix}.$$ From $$f(\phi(x,y)) = g(x)$$ it follows that $$0 = \frac{\partial(f\circ\phi)}{\partial y}(x,y) = f'(\phi(x,y))\binom{-\pi'(y)}{1} = -\frac{\partial f}{\partial x}(x-\pi(y),y)\cdot\pi'(y) + \frac{\partial f}{\partial y}(x-\pi(y),y).$$ Writing $$f_x$$ and $$f_y$$ for the partial derivatives then yields $$\pi'(y) = \frac{f_y(x-\pi(y),y)}{f_x(x-\pi(y),y)}.$$ By means of a similar reasoning one finds that $$g'(x) = f_x(x-\pi(y),y)$$. So, $$\pi'(y) = \frac{f_y(x-\pi(y),y)}{g'(x)}.$$