Continuity and ...

If $$x=t^{2},y=t^{3} \    then \ \  y_{2}=$$

Suppose $$f(x)=x+\cot x$$ then $$(\sin^{2}x)f''(x)+2x=$$

lf $$y=e^{-2x}\cos 3x$$ and $$\displaystyle \frac{d^{2}y}{dx^{2}}+a(\frac{dy}{dx})+by=0$$, then $$a=?,b=?$$

$$y=a\cos ec(b-x)$$, then $$yy_{2}-2y_{1}^{2}=$$

The derivative of $$f(\tan x)$$ w.r.t. $$g (\sec x)$$ at $$x=\displaystyle \frac{\pi }{4},$$ where $${f}'(1)=2$$ and $${g}'(\sqrt{2})=4,$$ is

If $$y=\displaystyle \frac{1}{1+x+x^{2}+x^{3}}$$, then $$y_{2}(0)=$$

If $$y^x=x^y$$, then find $$\displaystyle\frac{dy}{dx}$$.

If $$\displaystyle x=\frac{2t}{1+t^2}$$, $$\displaystyle y=\frac{1-t^2}{1+t^2}$$, then $$\displaystyle\frac{dy}{dx}$$ at $$t=2$$ is

$$\displaystyle\frac{dy}{dx}$$ for $$y=x^x$$ is

Let the function $$y=f(x)$$ be given by $$x=t^{5}-5t^{3}-20t+7$$ and $$y=4t^{3}-3t^{2}-18t+3$$, where $$t\epsilon \left ( -2, 2 \right )$$. Then $$f^{'}(x)$$ at $$t=1$$ is ?