Easy
Find the derivative: $\frac{d}{dx} \left[ \log \left( x + \frac{1}{x} \right) \right] = $
- A$x + \frac{1}{x}$
- B$\frac{1 + \frac{1}{x^2}}{1 + \frac{1}{x}}$
- C$\frac{1 - \frac{1}{x^2}}{x + \frac{1}{x}}$
- D$1 + \frac{1}{x}$
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Let $f:R \to R$ be such that $f(1) = 3$ and $f'(1) = 6$. Then $\lim_{x \to 0} \left\{ \frac{f(1 + x)}{f(1)} \right\}^{\frac{1}{x}}$ equals
DifficultIIT 2002
View SolutionIf $y=\log \left[\tan \sqrt{\frac{2^x-1}{2^x+1}}\right], x>0$, then $\left(\frac{d y}{d x}\right)_{x=1}=$
$x = \frac{1-\sqrt{y}}{1+\sqrt{y}} \Rightarrow \frac{dy}{dx}$ is equal to
Match the functions in List-$I$ with their derivatives given in List-$II$.
List-$I$ List-$II$ $A$. $\sec^{-1} x$ $I$. $\frac{1}{1-x^2}, x \in (-1, 1)$ $B$. $\tanh^{-1} x$ $II$. $\frac{-1}{|x| \sqrt{x^2+1}}, x \neq 0$ $C$. $\coth^{-1} x$ $III$. $\frac{1}{|x| \sqrt{x^2-1}}, |x| > 1$ $D$. $\operatorname{cosech}^{-1} x$ $IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$ $V$. $\frac{-1}{|x| \sqrt{1-x^2}}, |x| < 1, x \neq 0$
| List-$I$ | List-$II$ |
| $A$. $\sec^{-1} x$ | $I$. $\frac{1}{1-x^2}, x \in (-1, 1)$ |
| $B$. $\tanh^{-1} x$ | $II$. $\frac{-1}{|x| \sqrt{x^2+1}}, x \neq 0$ |
| $C$. $\coth^{-1} x$ | $III$. $\frac{1}{|x| \sqrt{x^2-1}}, |x| > 1$ |
| $D$. $\operatorname{cosech}^{-1} x$ | $IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$ |
| $V$. $\frac{-1}{|x| \sqrt{1-x^2}}, |x| < 1, x \neq 0$ |
$\frac{d}{d x}\left[\left(x^{\frac{5}{2}}-x^{\frac{3}{2}}+1\right)\left(x^2-3 x+5\right)\right]=$
DifficultTS EAMCET 2022
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