Right choice is (b) \(\frac{d^2 y}{dx^2}+(\frac{dy}{dx})^2\)=0
The best I can explain: Consider the function y=logx
Differentiating w.r.t x, we get
\(\frac{dy}{dx}=\frac{1}{x} \)–(1)
Differentiating (1) w.r.t x, we get
\(\frac{d^2 y}{dx^2}=-\frac{1}{x^2} \)
∴\(\frac{d^2 y}{dx^2}+(\frac{dy}{dx})^2=-\frac{1}{x^2}+(\frac{1}{x})^2\)
=-\(\frac{1}{x^2}+\frac{1}{x^2}\)=0.