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What is the partial derivative of $a_1x+b_1y$ with respect to $a_2x+b_2y$?

$$\frac\partial{\partial (a_2x+b_2y)}(a_1x+b_1y)$$It is zero?

Gaurav Gupta
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1 Answers1

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No. It is not, since the two expressions are not independent.

Say $f=a_2x+b_2y$ and $g=a_1x+b_1y$.

So, the problem now becomes, $$\frac{\partial g}{\partial f}$$ $$= \frac{\partial g}{\partial x}\cdot \frac{\partial x}{\partial f} + \frac{\partial g}{\partial y}\cdot \frac{\partial y}{\partial f}$$ $$= \frac{a_1}{a_2}+\frac{b_1}{b_2}$$

Hope this helps you.

SchrodingersCat
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  • Thanks for the answer. I have another question on this concept. can you please help me with https://math.stackexchange.com/questions/2294824/understanding-partial-derivatives-in-backpropagation-algorithm – Gaurav Gupta May 24 '17 at 11:43