Let $f=\frac{x^2w(y-z)t}{18l}$
then (imho):
$\frac{\partial{f(x,y,z)}}{\partial{w}}=\frac{x^2t(y-z)}{18l}$
However Wolfram Alpha produces a quite different result:
Wolfram alpha result
So who's wrong this time - me or the computer? If the latter, then why?
$\endgroup$ 24 Answers
$\begingroup$Wolfram is interpreting "$w(y-z)$" as a function of $y-z$, which clearly contains only y and z as variables. Try enclosing "$w$" in parentheses: $D[\frac{(x^2(w)(y-z)t)}{(18l)},w].$ It should work.
$\endgroup$ 1 $\begingroup$If you do $$\mathrm{D}\left[\frac{x^{2}w(y-z)t}{18l},w\right]$$ instead you will get the expected answer. Perhaps it doesn't parse "${2}w$" the way we might expect.
$\endgroup$ $\begingroup$I assume you meant taking the derivative with respect to $w$.
You are right, obviously, and Alpha is wrong. As for why... who knows? Submit a ticket with Wolfram. If you put in explicit multiplications, it works:
$\endgroup$ $\begingroup$When people set variables for programming, the variables are giving declarations of more than a letter many times so it is discriptive. For this reason, wolfram will read a concatenation of variables as one variable not two unless specified by parenthesis or multiplication.
If we had a times b times c, we couldn't enter abc since that will be treated as one variable not three.
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