Let $f:R \to R$ be such that $f(1) = 3$ and $f'(1) = 6$. Then $\lim_{x \to 0} \left\{ \frac{f(1 + x)}{f(1)} \right\}^{\frac{1}{x}}$ equals

If $y = \cos (\sin {x^2}),$ then at $x = \sqrt {\frac{\pi }{2}} , \frac{dy}{dx} = $

For $x > 0$ and $(x \log x) < 1$, if $y = \cot^{-1} \left( \frac{x - \log x}{x^2 \log_e x^2 + \log x^x} \right)$, then $\frac{dy}{dx} = \dots$

Find $\frac{dy}{dx}$ if $x-y=\pi$.

$\frac{d}{dx}(e^{x\sin x}) = $

Let $f(x)$ be a continuous and differentiable function for all real numbers. If $f(x + y) = f(x) - 3xy + f(y)$ and $\lim_{h \to 0} \frac{f(h)}{h} = 7$,then the value of $f'(x)$ is: