How do you differentiate $y =(sqrtx-3)^3$ using the chain rule?

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Answer 1 · 2018-07-06T14:21:33

$dy/dx=3/(2sqrtx) xx(sqrtx-3)^2$

$y=(sqrtx-3)^3$

Let
$t=sqrtx-3$

Let
$u=sqrtx$
$(du)/(dx)=1/(2sqrtx)$
$v=3$
$(dv)/(dx)=0$
$t=u-v$

$dt/dx=(du)/(dx)-(dv)/(dx)$

$dt/dx=1/(2sqrtx)-0$

$dt/dx=1/(2sqrtx)$

$y=t^3$

$dy/dx=3t^2dt/dx$

Substituting for t and dt/dx

$dy/dx=3xx(sqrtx-3)^2 xx 1/(2sqrtx)$

$dy/dx=3/(2sqrtx) xx(sqrtx-3)^2$

How do you differentiate y =(sqrtx-3)^3 y =(sqrtx-3)^3 using the chain rule?