Chain rule doubt

Question

I have a doubt of appling the chain rule. I have this $L$ function: $$ L = y\cdot log(\frac{e^{a x+b}}{e^{ax+b} + exp^{cx+d}}) $$ I can rewrite it as: $$ L = y\cdot log(p) $$ where $$ p = \frac{e^{v_{0}}}{e^{v_{0}}+e^{v_{1}}} $$ $$ v_{0} = ax+b $$ $$ v_{1} = cx+d $$

If I apply the chain rule I have this:

$$ \frac{\partial L}{\partial x} = \frac{\partial L}{\partial p} \cdot \frac{\partial p}{\partial v_{0}} \cdot \frac{\partial v_{0}}{\partial x} $$

But I know that I am missing somewhere the value of $\frac{\partial v_{1}}{\partial x}$

What I am doing wrong?

Answer 1

Your result is wrong because $p$ is a function of $x$ that contains the two functions $\nu_0$ and $\nu_1$ but not in the nested form $p(\nu_0(\nu_1))$. So we must write its derivative using also the quotient rule as:

$$ \frac{\partial p}{\partial x}=\frac{e^{\nu_0}\frac{\partial \nu_0}{\partial x}(e^{v_{0}}+e^{v_{1}})-e^{\nu_0}(e^{v_{0}}\frac{\partial \nu_0}{\partial x}+e^{v_{1}}\frac{\partial \nu_1}{\partial x})}{(e^{v_{0}}+e^{v_{1}})^2} $$

and the correct final result is $$ \frac{\partial L}{\partial x} = \frac{\partial L}{\partial p} \cdot \frac{e^{\nu_0}\frac{\partial \nu_0}{\partial x}(e^{v_{0}}+e^{v_{1}})-e^{\nu_0}(e^{v_{0}}\frac{\partial \nu_0}{\partial x}+e^{v_{1}}\frac{\partial \nu_1}{\partial x})}{(e^{v_{0}}+e^{v_{1}})^2} $$

Answer 2

Note that $$p=p(v_0,v_1)\\ \frac{dp}{dx}=\frac{\partial p}{\partial v_0}\frac{dv_0}{dx}+\frac{\partial p}{\partial v_1}\frac{dv_1}{dx}$$