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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.

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    $\begingroup$ If $f=1$ and $g=1$, $\pi$ can be anything. Hence, this is not a definition and $\pi$ need not be differentiable. $\endgroup$ – amsmath Oct 21 '19 at 21:46
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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, $$ \pi'(y) = \frac{f_y(x-\pi(y),y)}{g'(x)}. $$

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