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Solving: ddx(2(x2)−2(y2)=2x2y2)\frac{d}{dx}\left(2^{\left(x^2\right)}- 2^{\left(y^2\right)}=2x^2y^2\right)
Solve instead: dydx(2x2−2y2=2x2y2)\frac{dy}{dx}\left(2^{x^2}-2^{y^2}=2x^2y^2\right)

Exercise

dydx(2x2−2y2=2x2y2)\frac{dy}{dx}\left(2^{x^2}-2^{y^2}=2x^2y^2\right)

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Final answer to the exercise

y′=−ln⁡(2)⋅2(x2+1)x+4xy2−2y(ln⁡(2)2(y2)+2x2)y^{\prime}=\frac{-\ln\left(2\right)\cdot 2^{\left(x^2+1\right)}x+4xy^2}{-2y\left(\ln\left(2\right)2^{\left(y^2\right)}+2x^2\right)}

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dydx (2x2−2y2=2x2y2)
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