How do you use the chain rule to differentiate $y = {csc}^{3} (\frac{3 x}{2})$ $y=csc^3((3x)/2)$ ?

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How do you use the chain rule to differentiate $y = {csc}^{3} (\frac{3 x}{2})$ $y=csc^3((3x)/2)$ ?

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Answer 1 · 2018-01-30T09:48:50

$\frac{d}{d x} ({csc}^{3} (\frac{3 x}{2})) = - \frac{9}{2} \cdot {csc}^{3} (\frac{3 x}{2}) \cdot cot (\frac{3 x}{2})$ $d/dx( csc^3((3x)/2)) = -9/2 * csc^3((3x)/2) * cot((3x)/2)$

Explanation:

$\frac{d}{d x} {f (g (x))}^{n} = n \cdot {f (g (x))}^{n - 1} \cdot \frac{d}{d x} f (g (x)) \cdot \frac{d}{d x} g (x)$ $d/dx f(g(x))^n = n * f(g(x))^(n-1) * d/dx f(g(x)) * d/dx g(x)$

$\frac{d}{d x} csc (x) = - csc (x) cot (x)$ $d/dx csc(x) = -csc(x) cot(x)$

$\frac{d}{d x} \frac{3 x}{2} = \frac{3}{2}$ $d/dx (3x)/2 = 3/2$

Applying the first formula:

$\frac{d}{d x} ({csc}^{3} (\frac{3 x}{2})) = 3 \cdot {csc}^{2} (\frac{3 x}{2}) \cdot (- csc (\frac{3 x}{2}) \cdot cot (\frac{3 x}{2})) \cdot \frac{3}{2}$ $d/dx( csc^3((3x)/2)) = 3* csc^2((3x)/2)* (-csc((3x)/2)*cot((3x)/2))*3/2$

Simplifying:

$3 \cdot \frac{3}{2} = \frac{9}{2}$ $3*3/2 = 9/2$
${csc}^{2} (\frac{3 x}{2}) \cdot (- csc (\frac{3 x}{2})) = - {csc}^{3} (\frac{3 x}{2})$ $csc^2((3x)/2) * (-csc((3x)/2)) = -csc^3((3x)/2)$

We end with:

$\frac{d}{d x} ({csc}^{3} (\frac{3 x}{2})) = - \frac{9}{2} \cdot {csc}^{3} (\frac{3 x}{2}) \cdot cot (\frac{3 x}{2})$ $d/dx( csc^3((3x)/2)) = -9/2 * csc^3((3x)/2) * cot((3x)/2)$

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How do you use the chain rule to differentiate $y = {csc}^{3} (\frac{3 x}{2})$ $y=csc^3((3x)/2)$ ?

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