Question
Find $\frac{d y}{d x}$ if : $y^3+\cos (x y)=x^2-\sin (x+y)$
✓
Answer
Given that: $y^3+\cos (x y)=x^2-\sin (x+y)$
Differentiate $w . r . t . x$.
$
\begin{aligned}
& \frac{d}{d x}\left(y^3\right)+\frac{d}{d x}[\cos (x y)]=\frac{d}{d x}\left(x^2\right)-\frac{d}{d x}[\sin (x+y)] \\
& 3 y^2 \frac{d}{d x}(y)-\sin (x y) \frac{d}{d x}(x y)=2 x-\cos (x+y) \frac{d}{d x}(x+y) \\
& 3 y^2 \frac{d y}{d x}-\sin (x y)\left[x \frac{d y}{d x}+y(1)\right]=2 x-\cos (x+y)\left[1+\frac{d y}{d x}\right] \\
& 3 y^2 \frac{d y}{d x}-x \sin (x y) \frac{d y}{d x}-y \sin (x y)=2 x-\cos (x+y)-\cos (x+y) \frac{d y}{d x} \\
& 3 y^2 \frac{d y}{d x}-x \sin (x y) \frac{d y}{d x}+\cos (x+y) \frac{d y}{d x}=2 x+y \sin (x y)-\cos (x+y) \\
& {\left[3 y^2-x \sin (x y)+\cos (x+y)\right] \frac{d y}{d x}=2 x+y \sin (x y)-\cos (x+y) } \\
\therefore & \frac{d y}{d x}=\frac{2 x+y \sin (x y)-\cos (x+y)}{3 y^2-x \sin (x y)+\cos (x+y)}
\end{aligned}
$
Differentiate $w . r . t . x$.
$
\begin{aligned}
& \frac{d}{d x}\left(y^3\right)+\frac{d}{d x}[\cos (x y)]=\frac{d}{d x}\left(x^2\right)-\frac{d}{d x}[\sin (x+y)] \\
& 3 y^2 \frac{d}{d x}(y)-\sin (x y) \frac{d}{d x}(x y)=2 x-\cos (x+y) \frac{d}{d x}(x+y) \\
& 3 y^2 \frac{d y}{d x}-\sin (x y)\left[x \frac{d y}{d x}+y(1)\right]=2 x-\cos (x+y)\left[1+\frac{d y}{d x}\right] \\
& 3 y^2 \frac{d y}{d x}-x \sin (x y) \frac{d y}{d x}-y \sin (x y)=2 x-\cos (x+y)-\cos (x+y) \frac{d y}{d x} \\
& 3 y^2 \frac{d y}{d x}-x \sin (x y) \frac{d y}{d x}+\cos (x+y) \frac{d y}{d x}=2 x+y \sin (x y)-\cos (x+y) \\
& {\left[3 y^2-x \sin (x y)+\cos (x+y)\right] \frac{d y}{d x}=2 x+y \sin (x y)-\cos (x+y) } \\
\therefore & \frac{d y}{d x}=\frac{2 x+y \sin (x y)-\cos (x+y)}{3 y^2-x \sin (x y)+\cos (x+y)}
\end{aligned}
$
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