Differentiate F(x)=xe^xCSCx

Question

This is a Calculus Question. Please show work if possible

Answer 1

Hi Antonio,

 

Surendra's answer is correct, but I will try to show why.

 

First, csc(x) is the same thing as 1/sin(x), so I will replace it as such - I find the "standard" trig functions (i.e. sin, cos, etc) easier to deal with.

 

So, the function we are trying to take the derivative of is:

 

    (x*e^x)/(sin(x))

 

The quotient rule for derivatives states that, if we have two functions, one divided by the other ( f(x)/g(x) ), the derivative is equal to:

 

     d    f(x)    =  g(x)*f'(x) - f(x)*g'(x)

     dx  g(x)                 (g(x))²

 

In this case, the numerator is f(x) = x*e^x

and the denominator is g(x) = sin(x)

 

So, using the product rule, f'(x) = x*e^x + e^x

and g'(x) = cos(x)

 

The overall derivative is then:

 

    sin(x)*(x*e^x + e^x) - (x*e^x)(cos(x))

                   (sin(x))²

 

Distributing and simplifying:

 

    (x)(e^x)(sin(x)) + (e^x)(sin(x)) - (x)(e^x)(cos(x))

                         (sin(x))²

 

    (x)(e^x)(csc(x)) + (e^x)(csc(x)) - (x)(e^x)(csc(x))(cot(x))

 

    (e^x)(csc(x)) * (x + 1 - (x)(cot(x)))

 

 

Answer 2

x(-e^x * cosecx cotx + cosecx * e^x)+e^x * cosecx