How do you differentiate $y = {(\sqrt{x} - 3)}^{3}$ $y =(sqrtx-3)^3$ using the chain rule?

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How do you differentiate $y = {(\sqrt{x} - 3)}^{3}$ $y =(sqrtx-3)^3$ using the chain rule?

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Answer 1 · 2018-07-06T14:21:33

$\frac{d y}{d x} = \frac{3}{2 \sqrt{x}} \times {(\sqrt{x} - 3)}^{2}$ $dy/dx=3/(2sqrtx) xx(sqrtx-3)^2$

Explanation:

$y = {(\sqrt{x} - 3)}^{3}$ $y=(sqrtx-3)^3$

Let
$t = \sqrt{x} - 3$ $t=sqrtx-3$

Let
$u = \sqrt{x}$ $u=sqrtx$
$\frac{d u}{d x} = \frac{1}{2 \sqrt{x}}$ $(du)/(dx)=1/(2sqrtx)$
$v = 3$ $v=3$
$\frac{d v}{d x} = 0$ $(dv)/(dx)=0$
$t = u - v$ $t=u-v$

$\frac{d t}{d x} = \frac{d u}{d x} - \frac{d v}{d x}$ $dt/dx=(du)/(dx)-(dv)/(dx)$

$\frac{d t}{d x} = \frac{1}{2 \sqrt{x}} - 0$ $dt/dx=1/(2sqrtx)-0$

$\frac{d t}{d x} = \frac{1}{2 \sqrt{x}}$ $dt/dx=1/(2sqrtx)$

$y = t^{3}$ $y=t^3$

$\frac{d y}{d x} = 3 t^{2} \frac{d t}{d x}$ $dy/dx=3t^2dt/dx$

Substituting for t and dt/dx

$\frac{d y}{d x} = 3 \times {(\sqrt{x} - 3)}^{2} \times \frac{1}{2 \sqrt{x}}$ $dy/dx=3xx(sqrtx-3)^2 xx 1/(2sqrtx)$

$\frac{d y}{d x} = \frac{3}{2 \sqrt{x}} \times {(\sqrt{x} - 3)}^{2}$ $dy/dx=3/(2sqrtx) xx(sqrtx-3)^2$

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How do you differentiate $y = {(\sqrt{x} - 3)}^{3}$ $y =(sqrtx-3)^3$ using the chain rule?

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How do you differentiate $y = {(\sqrt{x} - 3)}^{3}$ $y =(sqrtx-3)^3$ using the chain rule?