Question

Can anyone say how this happened​

Answer 1

Here,

$\longrightarrow P'(x)=2y\,\dfrac{dy}{dx}$

Differentiating with respect to $x,$

$\longrightarrow P''(x)=\left(2y\,\dfrac{dy}{dx}\right)'$

By product rule,

$\longrightarrow P''(x)=(2y)'\,\dfrac{dy}{dx}+2y\left(\dfrac{dy}{dx}\right)'\quad\quad\dots(1)$

Here,

• $(2y)'=2y'=2\cdot\dfrac{dy}{dx}$

• $\left(\dfrac{dy}{dx}\right)'=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=\dfrac{d^2y}{dx^2}$

Then (1) becomes,

$\longrightarrow P''(x)=2\cdot\dfrac{dy}{dx}\cdot\dfrac{dy}{dx}+2y\,\dfrac{d^2y}{dx^2}$

$\longrightarrow P''(x)=2\left(\dfrac{dy}{dx}\right)^2+2y\,\dfrac{d^2y}{dx^2}$

Or,

$\longrightarrow P''(x)=2y\,\dfrac{d^2y}{dx^2}+2\left(\dfrac{dy}{dx}\right)^2$