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I am having trouble calculating partial derivatives of a simple function.

The function is:

$$ y(a,b,c)=\frac {0.99821*(a-b)}{c-b} $$

And I need to calculate $ \frac {\partial y}{\partial a} $, $\frac {\partial y}{\partial b}$ and $\frac {\partial y}{\partial c}$ .

If I am not wrong, to calculate $\frac {\partial y}{\partial a}$, I need to consider $b$ and $c$ as constants. However, derivatives of constants are always 0, thing that will make the denominator of the initial function equal with 0.

Where am I mistaking?

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Hold $b$ and $c$ constant to find the derivative with respect to $a$. Thus \begin{equation} \frac{\partial y}{\partial a} = \frac{0.99821}{c-b} \end{equation} Similarly \begin{align} \frac{\partial y}{\partial b} &= \frac{\partial}{\partial b}\left(\frac{0.99821(a-b)}{c-b}\right)\\ &= \frac{1}{(c-b)^{2}}\left(-0.99821(c-b)-(-0.99821(a-b)) \right) \\ &= \frac{0.99821}{(c-b)^{2}}\left(a-c\right) \end{align} Where the second line above is found by using the quotient rule. The derivative with respect to $c$ is found similarly.

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and last but not least $$\frac{\partial y}{\partial c}=- 0.99821\,{\frac {a-b}{ \left( c-b \right) ^{2}}}$$

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