Tag: differencial calculus - differenciability and methods of differnciation

If $u=e^{x}(xcosy-ysiny)$ then $\frac{d^{2}y}{dx^{2}}+\frac{d^{2}u}{dy^{2}}=0$.

If $U=tan^{-1}(\dfrac{x^3+y^3}{x+y})$ , then $x\dfrac{du}{dx}+y\dfrac{du}{dy}=sinu$.

If $y=\sqrt{x}-\dfrac{1}{\sqrt{x}}$, then $2x\dfrac{dy}{dx}+y$=

If $x\sqrt {1+y}+y\sqrt {1+x}=0$ then  $\dfrac {dy}{dx}=\dfrac {1}{(1+x)^{2}}$

Let $f(x)$ be a function continuous on $[1, 2]$ and differentiable on $(1, 2)$ satisfying $f(1)=2, f(2)=3$ and $f'(x)\ge 1\forall x\in (1, 2)$. Define $g(x)=\displaystyle \int _{1}^{x}{f(t)dt}\forall x\in [1, 2]$ then the greatest value of  $g(x)$ on $[1, 2]$ is-

Given $y = x \sqrt{x^2+1}, \dfrac{dy}{dx}$=

If $\sin { { y+e }^{ -x\cos { y }  } } =e\quad then\quad \frac { dy }{ dx } \quad at\quad (1,\pi )$ is equal to 

If $\displaystyle \cos^{-1}\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )=\log a$ then $\displaystyle \frac{dy}{dx}$ is equal to

If $\displaystyle y=\sec(\tan^{-1}x)$, then $\displaystyle \frac {dy}{dx}$ at $x=1$ is equal to

If $y=sec(tan^{-1}x)$, then $\displaystyle\frac{dy}{dx}$ is.