Determine a function $h$ $h$ such that $\frac{d}{d x} [cot (2 x) h (x)] = - 2 x cot (2 x) + 2 x^{2} {csc}^{2} (2 x)$ $d/(dx)[cot(2x)h(x)] = -2xcot(2x)+2x^2csc^2(2x)$ ?

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Determine a function $h$ $h$ such that $\frac{d}{d x} [cot (2 x) h (x)] = - 2 x cot (2 x) + 2 x^{2} {csc}^{2} (2 x)$ $d/(dx)[cot(2x)h(x)] = -2xcot(2x)+2x^2csc^2(2x)$ ?

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Answer 1 · 2017-03-16T01:52:07

$h (x) = - x^{2}$ $h(x) = - x^2$

Explanation:

$\frac{d}{d x} [cot (2 x) h (x)]$ $d/(dx)[cot(2x)h(x)]$ can be found using the product rule. We get

$\frac{d}{d x} [cot (2 x) h (x)] = \frac{d}{d x} (h (x)) \cdot cot (2 x) + h (x) \cdot \frac{d}{d x} (cot (2 x))$ $d/(dx)[cot(2x)h(x)] = d/dx(h(x)) * cot(2x) + h(x) * d/dx(cot(2x))$ #

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

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

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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Determine a function $h$ $h$ such that $\frac{d}{d x} [cot (2 x) h (x)] = - 2 x cot (2 x) + 2 x^{2} {csc}^{2} (2 x)$ $d/(dx)[cot(2x)h(x)] = -2xcot(2x)+2x^2csc^2(2x)$ ?

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Explanation:

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

1 Answer

Explanation:

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Impact of this question

Determine a function $h$ $h$ such that $\frac{d}{d x} [cot (2 x) h (x)] = - 2 x cot (2 x) + 2 x^{2} {csc}^{2} (2 x)$ $d/(dx)[cot(2x)h(x)] = -2xcot(2x)+2x^2csc^2(2x)$ ?