JEE Main 2018 — Differentiation Question with Solution

Given, x² + y² + sin y = 4

After differentiating the above equation  w.r.t.x  we get

2x + 2y $$\frac{dy}{dx}$$ + cos y $$\frac{dy}{dx}$$ = 0          . . . . (1)

$$\Rightarrow$$  2x + (2y + cos y) $$\frac{dy}{dx}$$ = 0

$$\Rightarrow$$  $$\frac{dy}{dx}$$ = $$\frac{-2x}{2y+\cos y}$$

At ($$-$$ 2, 0), $${\left(\frac{dy}{dx}\right)}_{\left(-2,0\right)}$$ = $$\frac{-2x-2}{2\times 0+\cos 0}$$

$$\Rightarrow$$  $${\left(\frac{dy}{dx}\right)}_{\left(-2,0\right)}$$ = $$\frac{4}{0+1}$$

$$\Rightarrow$$   $${\left(\frac{dy}{dx}\right)}_{\left(-2,0\right)}$$ = 4          . . . . .(2)

Again differentiating equation (1) w.r.t  to  x, we get

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

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

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

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

So, at ($$-$$ 2, 0),

$$\frac{d^{2}y}{d{x}^2}$$ = $$\frac{-2-\left(2-0\right)\times 4^2}{2\times 0+1}$$

$$\Rightarrow$$  $$\frac{d^{2}y}{d{x}^2}$$ = $$\frac{-2-2\times 16}{1}$$

$$\Rightarrow$$ $$\frac{d^{2}y}{d{x}^2}$$ = $$-$$ 34