Tag: methods of differentiation

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Questions Related to methods of differentiation

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The function u = e^x(x cos y - y sin y) is a standard form where the second partial derivatives satisfy the Laplace-like condition d^2u/dx^2 + d^2u/dy^2 = 0. The statement is true.

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

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

  1. $\sqrt{x}$
  2. $2\sqrt{x}$
  3. $3\sqrt{x}$
  4. $\dfrac{\sqrt{x}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$y=\sqrt x-\dfrac{1}{\sqrt x}\ \dfrac{dy}{dx}=\dfrac{d}{dx}(x)^{\dfrac{1}{2}}-\dfrac{d}{dx}(\dfrac{1}{\sqrt x})=\dfrac{1}{2\sqrt x}-(-\dfrac{1}{2})x^{-\dfrac{3}{2}}=\dfrac{1}{2\sqrt x}+\dfrac{x^{-\dfrac{3}{2}}}{2}\ \dfrac{2x dy}{dx}+y=2x[\dfrac{1}{2\sqrt x}+\dfrac{x^{-\dfrac{3}{2}}}{2}]+\sqrt x-\dfrac{1}{\sqrt x}\ \quad =\sqrt x+\dfrac{1}{\sqrt x}+\sqrt x-\dfrac{1}{\sqrt x}\ \quad=2\sqrt x$


Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

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-

  1. $3$
  2. $5$
  3. $\dfrac{5}{2}$
  4. $\dfrac{3}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since f'(x) >= 1, f(x) is strictly increasing. Given f(1)=2 and f(2)=3, the function g(x) = integral from 1 to x of f(t) dt is also increasing, so its maximum occurs at the upper bound x=2. The integral of f(t) from 1 to 2 is bounded by the trapezoid area (f(1)+f(2))/2 * (2-1) = 2.5.

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

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

  1. $ \sqrt{x^2+1}$
  2. $\dfrac{2x^2+1}{ \sqrt{x^2+1}}$
  3. $\dfrac{3x^2+1}{ \sqrt{x^2+1}}$
  4. $\dfrac{3x^2+2}{ \sqrt{x^2+1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$y=x\sqrt{x^2+1}$

$\Rightarrow \dfrac{d}{dx}y=\dfrac{d}{dx}x\sqrt{x^2+1}$

$=\dfrac{d}{dx}\left(x\right)\sqrt{x^2+1}+\dfrac{d}{dx}\left(\sqrt{x^2+1}\right)x$

$=1\cdot \sqrt{x^2+1}+\dfrac{x}{\sqrt{x^2+1}}x$

$=\dfrac{2x^2+1}{\sqrt{x^2+1}}$
Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

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

  1. $\sin { y } $
  2. $-x\cos { y } $
  3. $e$
  4. $\sin { y } -x\cos { y } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

$\sin y+e^{-x\cos y}=e$

$\cos y \dfrac{dy}{dx}+e^{-x\cos y}\left [ x\sin y \dfrac{dy}{dx}-\cos y \right ]=0$

$\Rightarrow \cos y \dfrac{dy}{dx}+e^{-x\cos y} x\sin y \dfrac{dy}{dx} -\cos ye^{-x\cos y}=0$

$\dfrac{dy}{dx}[\cos y+x\sin y e^{-x\cos y}]=\cos y e^{-x\cos y}$

$\dfrac{dy}{dx}=\dfrac{\cos y e^{-x\cos y}}{\cos y+x\sin y e^{-x\cos y}}$

$\left [ \dfrac{dy}{dx} \right ] _{(1, \pi )}=\dfrac{\cos \pi  e^{-\cos \pi}}{\cos \pi+1 \sin \pi e^{-\cos \pi }}$

$\left [ \dfrac{dy}{dx} \right ] _{(1, \pi )}=\dfrac{-1 \times e}{-1+0 \times e}=e$
Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

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

  1. $\dfrac{y}{x}$
  2. $\dfrac{x}{y}$
  3. $\displaystyle\dfrac{ x^{2}}{y^{2}}$
  4. $\displaystyle\dfrac{ y^{2}}{x^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\displaystyle \cos^{-1}\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )=\log a$ 
$\Rightarrow \displaystyle \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\cos \log a=A$ (say)
Putting $u=\dfrac{y}{x}$ and applying componendo and dividendo, we have
$\left ( \dfrac{y}{x} \right )^{2}=u^{2}=\left ( 1-A \right )\left ( 1+A \right )$
$\Rightarrow \dfrac{y}{x}=\sqrt{\left ( 1-A \right )\left ( 1+A \right )}\Rightarrow x\dfrac{dy}{dx}-y=0$
$\Rightarrow $   $\dfrac{dy}{dx}=\dfrac{y}{x}$
Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

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

  1. $\displaystyle \frac {1}{\sqrt {2}}$
  2. $\displaystyle \frac {1}{2}$
  3. $1$
  4. $\sqrt {2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $\displaystyle y=\sec(\tan^{-1}x)=\sqrt {1+x^{2}}$
$\Rightarrow \displaystyle \frac {dy}{dx}=\frac {x}{\sqrt {1+x^{2}}}$
$\displaystyle\therefore  \left.\begin{matrix}\frac {dy}{dx}\end{matrix}\right| _{x=1}=\frac {1}{\sqrt {2}}$

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

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

  1. $\displaystyle\frac{x}{\sqrt{1+x^2}}$
  2. $\displaystyle\frac{-x}{\sqrt{1+x^2}}$
  3. $\displaystyle\frac{x}{\sqrt{1-x^2}}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $y=sec(\tan^{-1}x)$
On differentiating w.r.t. $x,$ we get
$\displaystyle\frac{dy}{dx}=sec(\tan^{-1}x)\cdot \tan(\tan^{-1}x)\frac{1}{1+x^2}$
$=\displaystyle\frac{x}{1+x^2}\sqrt{1+x^2}$
$[\because \tan^{-1} x=sec^{-1}(\sqrt{1+x^2})]$
$=\displaystyle\frac{x}{\sqrt{1+x^2}}$