Question

Find the Derivative of csc2x/e to negative x

Answer 1

Is the question:
\[\frac{d}{dx}\frac{\csc(2x)}{e^{-x}}\]

Answer 2

sorry for the late reply and yes it is

Answer 3

Okay, well we can simplify as:
\[\frac{d}{dx}\frac{\csc(2x)}{e^{-x}}=\frac{d}{dx}\frac{1}{\sin(2x)e^{-x}}=\frac{d}{dx}[\sin(2x)e^{-x}]^{-1}=\frac{d}{dx}[f(x)g(x)]^{-1}\]
Now, we can evaluate this as:
\[\eqalign{
&=-[f(x)g(x)]^{-2}[f(x)g(x)]' \\
&=-\frac{1}{[f(x)g(x)]^2}\times[f(x)g'(x)+g'(x)f(x)] \\
&=-\frac{f(x)g'(x)+g'(x)f(x)}{f(x)^2g(x)^2}
}\]

Answer 4

I got e