MCQ
${d \over {dx}}\left[ {\log \left( {x + {1 \over x}} \right)} \right] = $
  • A
    $\left( {x + {1 \over x}} \right)$
  • B
    ${{\left( {1 + {1 \over {{x^2}}}} \right)} \over {\left( {1 + {1 \over x}} \right)}}$
  • ${{\left( {1 - {1 \over {{x^2}}}} \right)} \over {\left( {x + {1 \over x}} \right)}}$
  • D
    $\left( {1 + {1 \over x}} \right)$

Answer

Correct option: C.
${{\left( {1 - {1 \over {{x^2}}}} \right)} \over {\left( {x + {1 \over x}} \right)}}$
c
(c) $\frac{d}{{dx}}\left\{ {\log \left( {x + \frac{1}{x}} \right)} \right\} = \frac{1}{{x + \frac{1}{x}}} \times \frac{d}{{dx}}\left( {x + \frac{1}{x}} \right) $

$= \frac{1}{{\left( {x + \frac{1}{x}} \right)}}\left( {1 - \frac{1}{{{x^2}}}} \right)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $f : R \rightarrow R$ and $g : R \rightarrow R$ defined by $f(x) = 2x + 3$ and $g(x) = x^2 + 7,$ then the value of $x$ for which $f(g(x)) = 25$ is:
Let $\quad \overrightarrow{ a }=\alpha \hat{ i }+3 \hat{ j }-\hat{ k }, \overrightarrow{ b }=3 \hat{ i }-\beta \hat{ j }+4 \hat{ k } \quad$ and $\overrightarrow{ c }=\hat{ i }+2 \hat{ j }-2 \hat{ k }$ where $\alpha, \beta \in R$, be three vectors. If the projection of $\vec{a}$ on $\vec{c}$ is $\frac{10}{3}$ and $\overrightarrow{ b } \times \overrightarrow{ c }=-6 \hat{ i }+10 \hat{ j }+7 \hat{ k }$, then the value of $\alpha+\beta$ equal to
The matrix $\text{A}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$ is a:
$\sin−10 $ is equal to:
$\int\frac{1}{7+5\cos\text{x}}\text{ dx}=$
The equation of motion of a particle is $\text{s} = \text{2t}^2 + \sin\text{2t,}$ where $s$ is in metres and $t$ is in seconds. The velocity of the particle when its acceleration is $2m/\sec^2$, is:
If $y = x + e^x$ then $\frac{{{d^2}x}}{{d{y^2}}}$ is :
The area of the region bounded by the parabola $y=x^2+1$ and the staight line $x+y=3$ is given by:
If $\vec a \,$ and $\vec b \,$ are non-collinear vectors, then the value of $\alpha $ for which the vectors $\vec u  = \left( {\alpha  - 2} \right)\vec a \, + \vec b $ and $\,\vec v  = \left( {2 + 3\alpha } \right)\vec a \, - 3\vec b $ are collinear is :
Let $f (x)$ and $g (x)$ be two continuous functions defined from $R \rightarrow R$, such that $f (x_1) > f (x_2)$ and $g (x_1) < g (x_2), \forall x_1 > x_2$ , then solution set of $f\,\left( {\,g({\alpha ^2} - 2\alpha )\,} \right) >f\,\left( {\,g(3\alpha - 4)\,} \right)$ is