If the rate of change of x is 2 unit $s^-1$ , Find the rate of change of y in $y=3/((2x-3)^3$ when x=2?

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Answer 1 · 2018-01-31T17:22:50

$"The reqd. rate="-36" u/sec."$ .

$y=3/(2x-3)^3=3(2x-3)^-3$ .

Diff.ing w.r.t. $t$ using the Chain Rule, we have,

$dy/dt=d/dx{3(2x-3)^-3}*dx/dt$ ,

$={3(-3)(2x-3)^-4*d/dx(2x-3)}*dx/dt$ ,

$rArr dy/dt=-18(2x-3)^-4*dx/dt$ .

Here, $dx/dt"=the rate of change of "x=2" u/sec."$ ,

The reqd. rate of change of $y=dy/dt$ .

$:."The reqd. rate="[dy/dt]_(x=2)$ ,

$=[-18(2x-3)^-4]_(x=2)*(2)$ ,

$=-36{2(2)-3}^-4$ .

$rArr "The reqd. rate="-36" u/sec."$ .

If the rate of change of x is 2 unit s^-1s^-1, Find the rate of change of y in y=3/((2x-3)^3y=3/((2x-3)^3 when x=2?