Determine a function #h# such that#d/(dx)[cot(2x)h(x)] = -2xcot(2x)+2x^2csc^2(2x)#?

1 Answer
Jim H
Mar 16, 2017

#h(x) = - x^2#

Explanation:

#d/(dx)[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))##

# = 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#

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