Question

How do you differentiate `f(x)=sin2x * cot(5-x)` using the product rule?

Answer 1

`2 cos(2x) cot(5-2x)+sin (2x)cot(5-x)csc(5-x)`

Explanation:

The rule for differentiating a product of two functions is

`d/dx (uv) = (du)/dx v+u (dv)/dx`

Thus

`f(x)=sin2x * cot(5-x) qquad implies qquad`

`d/dx f(x)=d/dx (sin2x )* cot(5-x) + sin2x * d/dx(cot(5-x))`
`qquad = (2 cos(2x) )* cot(5-2x)+sin (2x)*(-cot(5-x)csc(5-x)times (-1))`
`qquad = 2 cos(2x) cot(5-2x)+sin (2x)cot(5-x)csc(5-x)`

where we have used the standard rules

`d/dx sin x = cos x, qquad d/dx cot x = -cot x csc x`

Answer 2

The derivative of `sin2xcot(5-x)` with respect to `x` is
`2cos2xcot(5-x)+sin2xcsc^2(5-x)`.

Explanation:

The product rule states

`(d[f(x)*g(x)])/(dx)=(df(x))/(dx)g(x)+f(x)(dg(x))/(dx)`

Here, `f(x)=sin2x`, and `g(x)=cot(5-x)`

Note that `f(x)` and `g(x)` are both COMPOSITE functions, so to solve this problem we will also need to use the chain rule which states that

`(df(g(x)))/(dx)=f^'(g(x))*g^'(x)`.

Now that we know the rules, let's compute the derivative

`(dsin2x*cot(5-x))/(dx)=(dsin2x)/(dx)cot(5-x)+sin2x(dcot(5-x))/(dx)`

`=2cos2xcot(5-x)+sin2x[-csc^2(5-x)*(-1)]`

`=2cos2xcot(5-x)+sin2xcsc^2(5-x)`