$\text{Differentiate }y=\sqrt{x}\text{ from the first principle}$

$\text{Differentiate }y=\tan x\text{ from the first principle}$

$\text{Differentiate }y=\text{ }x{{e}^{x}}\text{ from the first principle}$ 

$\begin{align}  & \text{Establish the formula for differentiating a quotient of two functions }y=\frac{u}{v}\text{ } \\ & \text{then }\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{{{v}^{2}}} \\\end{align}$

$\begin{align}  & \text{If }y=uvw\text{ } \\ & \text{show that }\frac{d}{dx}=u\frac{d}{dx}(vw)+v\frac{d}{dx}(uw)+w\frac{d}{dx}(uv) \\\end{align}$

$\text{Differentiate with respect to }x,\text{ }y=3x\sin x\cos x$

$\text{Differentiate }y=8{{x}^{2}}(1+\sin x)(1+\cos x)\text{ wrt }x$

$\text{Differentiate }y=\frac{1}{1+\cos x}\text{ with respect to }x$

$\text{Differentiate }y={{\sin }^{-1}}x\text{ with respect to }x$

$\text{Differentiate }y={{x}^{x}}+{{e}^{\tan x}}\text{ with respect to }x$

$\text{Differentiate }y={{\log }_{a}}{{x}^{2}}\text{ wrt }x$

$\text{Differentiate }y=x{{\sin }^{-1}}x\text{ wrt }x$

$\text{Differentiate }y={{e}^{\sin x}}\text{ wrt }x$

$\text{If }y=A{{e}^{mx}}+B{{e}^{-mx}},\text{ where }A,B,\text{ and }M\text{ are constant Show that }\frac{{{d}^{2}}y}{d{{x}^{2}}}={{m}^{2}}y$

$\text{If }x=a(\theta -\sin \theta )\text{ and }y=a(1-\cos \theta ),\text{ show that }1+{{\left( \frac{dy}{dx} \right)}^{2}}={{\operatorname{cosec}}^{2}}\frac{\theta }{2}$

$\text{Prove that }\frac{d}{dx}\left( \frac{1+\sin x+\cos x}{1-\sin x+\cos x} \right)=\frac{1}{1-\sin x}$

$\text{If }y={{\tan }^{-1}}x\text{ prove that }(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+2x\frac{dy}{dx}=0$

$\text{Differentiate with respect to }x,\text{ }y={{({{e}^{x}}\sin x)}^{2}}$