Let $f(x)$ and $g(x)$ be twice differentiable functions such that $f(x) = x^2 + g'(1)x + g''(2)$ and $g(x) = f(1)x^2 + xf'(x) + f''(x)$. Then $f(x) - g(x) =$

If $\sqrt{\frac{y}{x}}+4 \sqrt{\frac{x}{y}}=4$, then $\frac{d y}{d x}=$

Let $f$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1) .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int_{0}^{x} f(t) \,d t$ for $0 \leq x \leq 1$ and $f(0)=0$,then $\lim_{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} f(t) \,d t$ is:

For the curve $\sqrt{x} + \sqrt{y} = 1$,the value of $\frac{dy}{dx}$ at the point $\left( \frac{1}{4}, \frac{1}{4} \right)$ is:

If $\log(\sqrt{1+x^2}-x) = y(\sqrt{1+x^2})$,then $(1+x^2) \frac{dy}{dx} + xy =$

If $\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0$ and $x \neq y$,then $\frac{dy}{dx}$ is equal to: