Limits and Deri...

Differentiate the following w.r.t. $$x$$.
$$\sin^{2} \sqrt {x}$$.

$$\lim _{ x\rightarrow { 0 }^{ + } }{ \left( { \left( x\cos { x }  \right)  }^{ x }+{ \left( \cos { x }  \right)  }^{ \frac { 1 }{ \ln { x }  }  }+{ \left( x\sin { x }  \right)  }^{ x } \right)  } $$ is equal to

$$\dfrac{\displaystyle \lim_{h \rightarrow 0}(h+1)^2}{\displaystyle \lim_{h\rightarrow 0}(1+h)^{2/h}}$$ is equal to

$$\mathop {\lim}\limits_{x \to \frac{\pi}{2}} \tan x = $$

$$\lim _{ x\rightarrow 0 }{ \log _{ \left( \tan ^{ 2 }{ x }  \right)  }{ \left( \tan ^{ 2 }{ 2x }  \right) = }  }$$

$$\displaystyle \lim_{n \rightarrow \infty} {^{n}C_{c}}\left(\dfrac {m}{n}\right)^{x}\left(1-\dfrac {m}{n}\right)^{n-x}$$ equal to

$$\underset{h \rightarrow 0}{lim} \dfrac{\sqrt{x + h} -\sqrt{x}}{h}$$ is equal to 

Let $$y=a\cos t+b\sin t$$ then $$\dfrac{d^2y}{dt^2}=$$

The difference of slopes of lines represent by $${y^2} - 2xy{\sec ^2}\alpha  + \left( {3 + {{\tan }^2}\alpha } \right)\left( {{{\tan }^2}\alpha  - 1} \right){x^2} = 0$$ is

$$\lim\limits_{x\to 0}\dfrac{1-\cos x }{x^2}=$$