Question

निम्नलिखित प्रश्न में ${\displaystyle \frac{dy}{dx}}$ ज्ञात कीजिए:

sin² y + cos xy = k

Answer 1

चूँकि, sin² y + cos xy = k

दोनों पक्षों का x के साक्षेप अवकलन करने पर,

${\displaystyle \frac{d}{dx}\left({\sin }^{2}y\right)+\frac{d}{dx}\left(\cos xy\right)=\frac{d}{dx}\left(k\right)}$

${\displaystyle \Rightarrow 2\sin y\cos y\frac{dy}{dx}+(-\sin xy)\frac{d}{dx}\left(xy\right)=0}$

${\displaystyle \Rightarrow 2\sin y\cos y\frac{dy}{dx}-\sin xy[x\frac{dy}{dx}+y\frac{d}{dx}\left(x\right)]=0}$

${\displaystyle \Rightarrow 2\sin y\cos y\frac{dy}{dx}-x\sin xy\frac{dy}{dx}-y\sin xy=0}$

${\displaystyle \Rightarrow \sin 2y-x\sin xy\frac{dy}{dx}-y\sin xy=0}$

${\displaystyle \Rightarrow \frac{dy}{dx}(\sin 2y-x\sin xy)=y\sin xy}$

${\displaystyle \frac{dy}{dx}=\frac{y\sin xy}{(\sin 2y-x\sin xy)}}$