Question

Find the second derivative at the point (1,2), given the function below. y+3y^3-2x^2=10x+14

Answer 1

Step-by-step explanation:

$y + 3 {y}^{3} - 2 {x}^{2} = 10x + 14 \\ \therefore \: y + 3 {y}^{3} = 2 {x}^{2} + 10x + 14\\ differentiatig \: w.r.t. \: x \: on \: both \: sides : \\ \\ \frac{dy}{dx} + 3 \times 3 {y}^{2} \frac{dy}{dx} = 2 \times 2x + 10 + 0 \\ \\ \therefore \: \frac{dy}{dx} + 9{y}^{2} \frac{dy}{dx} = 4x + 10 \\ \\ differentiatig \: again \: w.r.t. \: x \: on \: \\ both \: sides : \\ \frac{d}{dx}(\frac{dy}{dx}) + 9 \times 2{y} \frac{d}{dx}(\frac{dy}{dx}) \: = 4 + 0 \\ \\ \therefore \:\frac{d^{2} y}{dx^{2}} + 18{y} \frac{d^{2}y}{dx^{2}}= 4 \\ \\ \therefore \:(1 + 18{y} )\frac{d^{2}y}{dx^{2}}= 4 \\ \\ \therefore \:\frac{d^{2}y}{dx^{2}} = \frac{4}{1 + 18{y} } \\ \\ \therefore \:[\frac{d^{2}y}{dx^{2}}] _{(1, \: \: 2)}= \frac{4}{1 + 18 \times 2} \\ \\ \therefore \:[\frac{d^{2}y}{dx^{2}}] _{(1, \: \: 2)}= \frac{4}{37}$