If \( y^2=ax+b,a,b\in\mathrm{R} \) where \( \mathrm{R} \) is the set of real numbers , then \( y\displaystyle\frac{d^2y}{dx^2}+\left(\displaystyle\frac{dy}{dx}\right)^2 \) is equal to :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Differentiating \( y^2=ax+b \) implicitly with respect to x,
\( \displaystyle 2y\frac{dy}{dx}=a \)
Differentiating again, with the product rule on the left and a constant on the right,
\( \displaystyle 2\left(\frac{dy}{dx}\right)^2+2y\frac{d^2y}{dx^2}=0 \)
\( \displaystyle y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=0 \)
Shortcut: The given expression is \( \frac{1}{2}\frac{d^2}{dx^2}(y^2) \), and \( y^2=ax+b \) is linear in x, so its second derivative is 0.