$\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\frac{{{e^{{e^{{x^2}}}}} - e}}{x}} \right)$ is

Consider $f(x) = \begin{cases} \tan^{-1}(\frac{\alpha x + \beta}{\gamma}) & x \in (0, \frac{1}{2}) \\ 0 & x = \frac{1}{2} \\ \ln(\beta x^2 + 2) & x \in (\frac{1}{2}, 1) \end{cases}$. If $f(x)$ is continuous and differentiable in its domain,then the value of $\alpha + \beta + \gamma$ is:

If $f(x)=x^2+g^{\prime}(1) x+g^{\prime \prime}(2)$ and $g(x)=f(1) x^2+x f^{\prime}(x)+f^{\prime \prime}(x),$ then the value of $f(4)-g(4)$ is equal to $...........$.

Let $f$ be a differentiable function and the equation of the normal to the graph of $y = f(x)$ at $x = 3$ is $3y = x + 18$. If $L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) - f\left( {3 + {{\left( {f\left( 3 \right) - x - 6} \right)}^2}} \right)}}{{{{\sin }^2}\left( {x - 1} \right)}}$,then:

Let $f: R \to R$ be a differentiable function such that $f(2) = 6$ and $f'(2) = \frac{1}{48}.$ Then $\lim_{x \to 2} \int_{6}^{f(x)} \frac{4t^3}{x - 2} dt$ equals

If $f:R \to R$ and $f(x)$ is a polynomial function of degree $10$ such that $f(x)=0$ has all real and distinct roots,then the equation $(f'(x))^2 - f(x)f''(x) = 0$ has: