I am trying to find the derivative of $f(x)= xe^x \csc x$, and I am not too sure how to even start.
Is it two terms or three? $xe^x$ and $\csc x$ or is it $x$, $e^x$ and $\csc x$? I can't get a proper answer either way.
With two terms I get $xe^x(-\csc x\cot x) + \csc x(e^x)$.
Best Answer
To find the derivative of $(abc)'$ you use repeated application of the product rule: $$ (abc)' = (ab)'c+abc' = (ab'+a'b)c+abc' = a'bc+ab'c+abc'. $$ In your case $a(x) = x$, $b(x) = \mathrm e^x$ and $c(x) = \operatorname{csc}(x)$, so $$ a' = 1, b' = \mathrm e^x \text{ and }c' = -\cot x\csc x. $$
To make it more clear: in $x \mathrm e^x\csc x$ you have three function rather than two, but $x\mathrm e^x$ is also a product of two functions, so $$ (x\mathrm e^x\csc x)' = (x\mathrm e^x)'\csc x+x\mathrm e^x(\csc x)'. $$ We can calculate the latter term, but what about $(x\mathrm e^x)'$? You again apply the product rule: $$ (x\mathrm e^x)' = x'\mathrm e^x+x(\mathrm e^x)'. $$