Question

Can anyone help me differentiate this function?

`y=xe^x cscx`

Answer 1

`f'(x)=e^xcsc(x)+xe^xcsc(x)-xe^xcsc(x)cot(x)`

Explanation:

For a function in the form

`f(x)=(fgh)`,

`f'(x)=(fgh)'=f'gh+fg'h+fgh'`

Here, we can say `f(x)=x, g(x)=e^x, h(x)=csc(x)`

Therefore,

`f'(x)=(d/dx(x))e^xcsc(x)+x(d/dx(e^x))csc(x)+xe^xd/dxcsc(x)`

`d/dx(x)=1`
`d/dxe^x=e^x` (The derivative of the natural exponential function is equal to the natural exponential function)
`d/dxcsc(x)=-csc(x)cot(x)`

Plugging in, we get:

`f'(x)=e^xcsc(x)+xe^xcsc(x)-xe^xcsc(x)cot(x)`