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Question Number 117486 by Canovas last updated on 12/Oct/20

Answered by john santu last updated on 12/Oct/20

$$\frac{dy}{dx}=2x.\left(2^y\right)+x^2.\ln \left(2\right).2^y\frac{dy}{dx}$$$$(1-\ln \left(2\right)x^2.2^y)\frac{dy}{dx}=2x.2^y$$$$\frac{dy}{dx}=\frac{x.2^{y+1}}{1-y.\ln \left(2\right)}.$$

Commented by Canovas last updated on 12/Oct/20

$$Thankssir!$$

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