1
$\begingroup$

how do I find $\frac{\partial q}{\partial k}$ of $q(k,l,m) = k\,p(k,l) + m^2$ ?

I have tried

$\frac{\partial q}{\partial k}= p(k,l) \times\begin{bmatrix}\frac{dk}{dm}+\frac{dl}{dm}\end{bmatrix} + p(k,l)\\ \frac{\partial q}{\partial l} = k\,p'_l(k,l)\, \\\frac{\partial q}{\partial m} = 2m$

$\endgroup$
2
$\begingroup$

Unless I am missing something here:

-ignore the $m^{2}$ because it is constant w.r.t. k

-Use the product rule on $kp(k,l)$ :

$\partial_{k}$$kp(k,l)$=$p(k,l)+k\partial_{k}p(k,l)$

I am not sure whether this is what you were asking though....

$\endgroup$
1
$\begingroup$

here it is: $$ \frac{\partial q}{\partial k}= 1p(k,l) + k \times\frac{\partial p}{\partial k}\\ \frac{\partial q}{\partial l}= k \times\frac{\partial p}{\partial l}\ \ \\\frac{\partial q}{\partial m} = \frac{\partial k}{\partial m}p(k,l) + k \times\frac{\partial p}{\partial m}+2m=\frac{\partial k}{\partial m}p(k,l)+k \times\begin{bmatrix}\frac{\partial p}{\partial k}\frac{\partial k}{\partial m}+\frac{\partial p}{\partial l}\frac{\partial l}{\partial m}\end{bmatrix} + 2m\ $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.