How do you use chain rule to find the derivative of $y = 2 {csc}^{3} (\sqrt{x})$ $y = 2 csc^3(sqrt(x))$ ?

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How do you use chain rule to find the derivative of $y = 2 {csc}^{3} (\sqrt{x})$ $y = 2 csc^3(sqrt(x))$ ?

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Answer 1 · 2015-03-01T01:13:17

You can easily illustrate the chain rule using Leibniz notation

Let $y = 2 u^{3}$ $y=2u^3$ then $\frac{d y}{d u} = 6 u^{2}$ $dy/(du)=6u^2$

Let $u = csc (w)$ $u=csc(w)$ then $\frac{d u}{d w} = - csc (w) cot (w)$ $(du)/(dw)=-csc(w)cot(w)$

Let $w = \sqrt{x}$ $w=sqrt(x)$ then $\frac{d w}{d x} = \frac{1}{2 \sqrt{x}}$ $(dw)/(dx)=1/(2sqrt(x))$

Now the chain rule is

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d w} \frac{d w}{d x}$ $dy/dx=(dy)/(du)(du)/(dw)(dw)/(dx)$

$\frac{d y}{d x} = 6 u^{2} (- csc (w) cot (w)) \frac{1}{2 \sqrt{x}}$ $dy/dx=6u^2(-csc(w)cot(w))1/(2sqrt(x))$

Remember that $u = csc (w)$ $u=csc(w)$ and $w = \sqrt{x}$ $w=sqrt(x)$

$\frac{d y}{d x} = 6 {csc}^{2} (\sqrt{x}) (- csc (\sqrt{x}) cot (\sqrt{x})) (\frac{1}{2 \sqrt{x}})$ $dy/dx=6csc^2(sqrt(x))(-csc(sqrt(x))cot(sqrt(x)))(1/(2sqrt(x)))$

Some simplifying

$\frac{d y}{d x} = \frac{- 3 {csc}^{3} (\sqrt{x}) cot (\sqrt{x})}{\sqrt{x}}$ $dy/dx=(-3csc^3(sqrt(x))cot(sqrt(x)))/sqrt(x)$

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How do you use chain rule to find the derivative of $y = 2 {csc}^{3} (\sqrt{x})$ $y = 2 csc^3(sqrt(x))$ ?

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How do you use chain rule to find the derivative of $y = 2 {csc}^{3} (\sqrt{x})$ $y = 2 csc^3(sqrt(x))$ ?