Basics of Differentiation

Suppose $f\left(x\right)=\tan \left({\sin }^{-1}\left(2x\right)\right)$, then $f^{\text{'}}\left(\frac{1}{4}\right)$ is equal to

On differentiating  $\sqrt{\tan x}$ with respect to $x$ from first principle, the solution would be:

For the differentiable function $f:\mathrm{ℝ}-\left\{0\right\}-\mathrm{ℝ},$ let $3f\left(x\right)+2f\left(\frac{1}{x}\right)=\frac{1}{x}-10,$ then $\left|f\left(3\right)+f^{\text{'}}\left(\frac{1}{4}\right)\right|$ is equal to

Let $f\left(x\right)=\frac{\sin x+\cos -\sqrt{2}}{\sin x-\cos x},x\in \left[0, \pi \right]-\left\{\frac{\pi }{4}\right\}$, then $f\left(\frac{7\pi }{12}\right)f^{"}\left(\frac{7\pi }{12}\right)$ is equal to

If $y=\frac{\cos 6x+6\cos 4x+15\cos 2x+10}{\cos 5x+5\cos 3x+10\cos x}$, then $\frac{dy}{dx}=$

If $x_1, x_2, x_3, ..., x_{n-1}$ be zeros of polynomial $P\left(x\right)=x^n+\alpha x+\beta$ where $x_i\ne x_j \forall i, j=1, 2, 3, ..., n-1$. The value of $Q\left(x\right)=\left(x_1-x_2\right)\left(x_1-x_3\right)....\left(x_1-x_{n-1}\right)$ is

If $\frac{{\cos }^4\alpha }{x}+\frac{{\sin }^4\alpha }{y}=\frac{1}{x+y}$ then $\frac{dy}{dx}=$

If $y={\cos }^{-1}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)$, then $\frac{dy}{dx}$ is

If $f\left(x\right)=\left[\begin{array}{ll}x{\tan }^{-1}x+{\sec }^{-1}\left(\frac{1}{x}\right),& x\in \left(-1,1\right)-\left\{0\right\}\\ \frac{\pi }{2}& \mathrm{if} x=0\end{array}\right.$, then $f^{\text{'}}\left(0\right)$ is

If $y=\log x+{\log }_{5}x+{\log }_{5}5$, then $\frac{dy}{dx}$ is

If $f\left(x\right)$ is a polynomial such that $f\left(x+1\right)=f\left(x\right)+2x \forall x\in R$ and $f\left(0\right)=1$ then $f^{\text{'}}\left(5\right)=$

Differentiate $arc\tan \left(\frac{2+x}{2-x}\right)$

The differentiation of  $y=7x^4$ w.r.t. $x$ is

If $f\left(2\right)=2, f\text{'}\left(2\right)=1$ then $\underset{x\to 2}{\lim }\frac{x f\left(2\right)-2f\left(x\right)}{x-2}=$

If $f\left(x\right)=1+x+x^2+\dots \dots +x^{1000}$, then $f^{\text{'}}(-1)=\dots \dots .$

$\frac{\mathrm{d}}{\mathrm{dx}}\left(3\cos \left(\frac{\pi }{6}+\mathrm{x}°\right)-4{\cos }^3\left(\frac{\pi }{6}+\mathrm{x}°\right)\right)=\_\_\_\_\_$

If the derivative of $f\left(x\right)=4\sqrt{x}-2$ with respect to $x$ is $f\text{'}\left(x\right)=\frac{k}{\sqrt{x}}$. Find the value of $k$?