Help with differentiation?

Question

Differentiate the following with respect to x:

3x / (tanx)^2

Answer 1

f(x) = 3x / tan^2x.

first, use the quotient rule:

f'(x) = [ (tan^2x)*3 - 3x* d (tan^2x)/dx ] / tan^4x

now, use the chain rule to differentiate tan^2x:

f'(x) = [ 3*tan^2x - 6x*tanx*sec^2x ] / tan^4x

      = 3*sec^2x [ sin^2x - 2x*tanx ] / tan^4x

      = 3 [ sin^2x - 2x*tanx ] / (tanx*sinx)^2.

Answer 2

Quotient rule:

y = u/v

u = 3x so u ' = 3

v = tan²(x) so v ' = 2tan(x)sec²(x)

y ' = (u ' v - v ' u) / v²

= (3tan²(x) - 6xtan(x)sec²(x)) / tan⁴(x)

= 3(tan(x) - 2xsec²(x)) / tan³(x)

OR:

y = 3x / tan²(x)

y = 3xcot²(x)

Now using the product rule

u = 3x so u ' = 3

v = cot²(x)  so v ' = -2cot(x)csc²(x)

y ' = u ' v + v ' u

y ' = 3cot²(x) - 6xcot(x)csc²(x)

y ' = 3cot(x)(cot(x) - 2xcsc²(x))

NOTE:

The person above is also correct