Given that $\frac{d}{dx}f(x) = f'(x)$. The relationship $f'(a + b) = f'(a) + f'(b)$ is valid if $f(x)$ is equal to

If $f(x)=x^{n}$, where $n$ is a non-negative integer, then the values of $n$ for which $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$ for all $\alpha, \beta > 0$ is:

Let $f$ and $g$ be differentiable functions satisfying $g'(a) = 2$,$g(a) = b$,and $f \circ g = I$ (identity function). Then $f'(b)$ is equal to

Let $f(x)$ be a differentiable function for all $x \in R$ and $f(x+y)=f(x)+f(y)-3xy$. If $\lim _{h \rightarrow 0} \frac{f(h)}{h}=7$, then $f^{\prime}(x)=$

$\frac{d}{dx} \left[ \left( \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} \right) \cot 3x \right] =$

If $\frac{d}{dx} \left( \frac{x^4 + x^2 + 1}{x^2 + x + 1} \right) = ax + b$,then $a - b =$