Problems in Partial Derivatives

Question

I'm trying to compute the Partial Derivatives for the equation:

$$k(a, b) = a^2 \exp\left(-\frac{1}{2b^2} (x-y)^2\right)$$ w.r.t. $b$

$x$ and $y$ are known; both $a$ and $b$ are hyper-parameters.

I'm a little confused on this problem...

Answer 1

If it is just to compute the partial derivative w.r.t to b then do as follows

$$k(a, b) = a^2 \exp\left(-\frac{1}{2b^2} (x-y)^2\right) = a^2\exp\left(-\frac{1}{2b^2}f(x,y)\right)$$

since the only place where $b$ occurs is in the exponent we can take the derivative as $$ \frac{\partial k}{\partial b} = a^2 \frac{\partial}{\partial b}\exp\left(-\frac{1}{2b^2}f(x,y)\right) $$ to take the derivtive of the exponent we proceed as $$ \frac{\partial}{\partial b}\exp\left(-\frac{1}{2b^2}f(x,y)\right) = \exp\left(-\frac{1}{2b^2}f(x,y)\right)\frac{\partial}{\partial b}\left(-\frac{1}{2b^2}f(x,y)\right)\\ =\exp\left(-\frac{1}{2b^2}f(x,y)\right) f(x,y)\frac{\partial}{\partial b}\left(-\frac{1}{2b^2}\right)\\ =\exp\left(-\frac{1}{2b^2}f(x,y)\right) f(x,y)\left(-\frac{-2}{2b^3}\right) $$ which leads to $$ \frac{\partial k}{\partial b} = \frac{a^2}{b^3}(x-y)^2\exp\left(-\frac{1}{2b^2}(x-y)^2\right) $$