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Step-by-step Solution

Find the derivative of $z=x\ln\left(x^2+y^2\right)=t^2y=t^{-2}$

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Step-by-step explanation

Problem to solve:

$\frac{d}{dt}\left(z=\left(\ln\:\left(x^2+y^2\right)\right)x=\left(t^2\right)y=\left(t^{-2}\right)\right)$

Answer

No steps currently available for this problem.
$\frac{d}{dt}\left(z=\left(\ln\:\left(x^2+y^2\right)\right)x=\left(t^2\right)y=\left(t^{-2}\right)\right)$

Main topic:

Differential calculus

Used formulas:

1. See formulas

Time to solve it:

~ 0.84 seconds

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