Download App

Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $x^y=e^{x-y}$ then $\frac{\text{dy}}{\text{dx}}$ is :
  • A
    $\frac{1+\text{x}}{1+\log\text{x}}$
  • B
    $\frac{1-\log\text{x}}{1+\log\text{x}}$
  • C
    Not defined.
  • ✓
    $\frac{\log\text{x}}{(1+\log\text{x})^2}$
✓

Answer

Correct option: D.
$\frac{\log\text{x}}{(1+\log\text{x})^2}$
We have, $x^y=e^{x-y}$
Taking $\log$ on both sides we get,
$\Rightarrow\text{y}\log\text{x}=(\text{x}-\text{y})\log)_\text{e}\text{e}$
$\Rightarrow\text{y}\log\text{x}=\text{x}-\text{y}$
$\Rightarrow\text{y}\log\text{x}+\text{y}=\text{x}$
$\Rightarrow\text{y}(1+\log\text{x})=\text{x}$
$\Rightarrow\text{y}=\frac{\text{x}}{(1+\log\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{x})\times1-\text{x}\times\Big(1+\frac{1}{\text{x}}\Big)}{(1+\log\text{x})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\log\text{x}-1}{(1+\log\text{x})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\log\text{x}}{(1+\log\text{x})^2}$

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

Explore more

More "M.C.Q (1 Marks)" from Continuity and Differentiability→Continuity and Differentiability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If A and B are two events such that P(A) = 0.4, P(B) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ then $\text{P}(\overline{\text{B}}\cap\text{A})$ equals.
→
The value of objective function is maximum under linear constraints
→
Let $R$ be the relation in the set $N$ given by $R =\{(a,\, b)\,:\, a=b-2,\, b>6\} .$ Choose the correct answer.
→
In Graphical solution the feasible solution is any solution to a LPP which satisfies.
→
If f(x) = |x - 3| and g(x) = fof(x), then for x > 10, g'(x) is equal to:
→
The angle between the lines $\frac{{x + 4}}{1} = \frac{{y - 3}}{2} = \frac{{z + 2}}{3}$ and $\frac{x}{3} = \frac{{y - 1}}{{ - 2}} = \frac{z}{1}$ is
→
Simplest form of
$\tan ^{-1}\left(\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right)$,$\pi$< x<$\frac{3 \pi}{2}$ is
→
The area of the region bounded by the curve $\text{y}=\sqrt{16-\text{x}^2}$ and $x-$axis is:
→
The value of the integral $\int\limits_0^{0.9} {[ - 2[x]]\,dx,} $ where $[.]$ denotes the greatest integer function is
→
Solve for $x :\{\text{x}\cos(\cot^{-1}\text{x})+\sin(\cot^{-1}\text{x})\}^2=\frac{51}{50}$
→