Determine a function hh such thatd/(dx)[cot(2x)h(x)] = -2xcot(2x)+2x^2csc^2(2x)ddx[cot(2x)h(x)]=2xcot(2x)+2x2csc2(2x)?

1 Answer
Mar 16, 2017

h(x) = - x^2h(x)=x2

Explanation:

d/(dx)[cot(2x)h(x)]ddx[cot(2x)h(x)] can be found using the product rule. We get

d/(dx)[cot(2x)h(x)] = d/dx(h(x)) * cot(2x) + h(x) * d/dx(cot(2x))ddx[cot(2x)h(x)]=ddx(h(x))cot(2x)+h(x)ddx(cot(2x))#

= h'(x) * cot(2x) +h(x) * (-2csc^2(2x))

= h'(x) * cot(2x) -2 h(x) * csc^2(2x)

We want this to be equal to = -2xcot(2x)+2x^2csc^2(2x),

So we must have -2h(x) = 2x^2,

so h(x) = -x^2