Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

I have tried

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

share|cite|improve this question
up vote 1 down vote accepted

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

share|cite|improve this answer

Unless I am missing something here:

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

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


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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