How do you find the derivatives of $y=ln(x^2y)$ ?

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Answer 1 · 2017-08-19T16:50:51

Please see below.

To find the derivatives of $x$ and $y$ with respect to some variable $t$ , use the chain rule (implicit differentiation).

$y = ln(x^2y)$ I would rewrite using properties of logarithms.

$y = 2lnx+lny$

Differentiate w.r.t. $t$

$dy/dt = 2/x dx/dt + 1/y dy/dt$

Solving for $dy/dt$

$(1-1/y)dy/dt = 2/x dx/dt$

$(y-1)/y dy/dt = 2/x dx/dt$

$dy/dt = (2y)/(x(y-1)) dx/dt$

And solving for $dx/dt$ ,

$(1-1/y)dy/dt = 2/x dx/dt$

$dx/dt = x/2(1-1/y) dy/dt$

$= (x(y-1))/(2y) dy/dt$

How do you find the derivatives of y=ln(x^2y)y=ln(x^2y)?