Question Number 161311 by CM last updated on 15/Dec/21 | ||
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$${Differentiate}\:{y}=\mathrm{sin}\:{xy} \\ $$ | ||
Answered by mr W last updated on 15/Dec/21 | ||
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$$\frac{{dy}}{{dx}}=\mathrm{cos}\:\left({xy}\right)\left({y}+{x}\frac{{dy}}{{dx}}\right) \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{y}\:\mathrm{cos}\:\left({xy}\right)}{\mathrm{1}−{x}\:\mathrm{cos}\:\left({xy}\right)} \\ $$ | ||
Commented by CM last updated on 16/Dec/21 | ||
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$${Thank}\:{you}\:{sir} \\ $$ | ||
Answered by 1549442205PVT last updated on 16/Dec/21 | ||
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$${we}\:{assume}\:{that}\:{y}\:{is}\:{a}\:{function}\:{of}\:{x}.{Then} \\ $$$${y}'={cosxy}.\left({y}+{xy}'\right)\Rightarrow{y}'\left(\mathrm{1}−{xcosxy}\right)={ycosxy} \\ $$$$\Rightarrow{y}'=\frac{{ycosxy}}{\mathrm{1}−{xcosxy}} \\ $$ | ||