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#