Mathematics

Differentiation and Algebraic Fractions

43 Questions

Differentiation combined with algebraic fractions involves solving implicit functions and evaluating complex mathematical ratios. Test items require calculating second derivatives, inverse trigonometric functions, and variable fraction equivalencies. This specialized mathematics topic appears regularly in high level aptitude screenings.

Implicit function derivativesAlgebraic fraction ratiosSecond derivative calculationInverse trigonometric differentiationVariable fraction equivalencies

Differentiation and Algebraic Fractions Questions

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

$\dfrac {d}{dx}(x^{\ell n x})$ is equal to

  1. $2x^{\ell n x-1}\ell n x$
  2. $x^{\ell n x-1}$
  3. $2/3(\ell n x)$
  4. $x^{\ell n x-1}.\ell n x$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $lnx=u$

$\therefore\ x={e}^{u}.$
$\frac { d }{ dx } \left( { x }^{ lmx } \right) =\frac { d }{ dx } \left( { e }^{ u.lnu } \right) $
$=\frac { d }{ dx } \left( { e }^{ { lnu }^{ 2 } } \right) =\frac { d }{ dx } \left( { u }^{ 2 } \right) $
$=2u.\frac { du }{ dx } $
$=2u.\frac { d }{ dx } \left( lnx \right) =\boxed{2lnx\quad { x }^{ lnx-1 }}$

Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

The simplified form of the expression given below is :$\dfrac{\dfrac{y^4-x^4}{x(x+y)}-\dfrac{y^3}{x}}{y^2-xy+x^2}$

  1. $1$
  2. $0$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given expression $\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }$ can be simplified as follows:

 
$\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { ({ y }^{ 2 })^{ 2 }-({ x }^{ 2 })^{ 2 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { ({ y }^{ 2 }-{ x }^{ 2 })({ y }^{ 2 }+{ x }^{ 2 }) }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \quad \quad \quad \quad \quad \quad \quad \left( \because \quad a^{ 2 }-b^{ 2 }=(a+b)(a-b) \right)$
$=\dfrac { \dfrac { ({ y }+x)(y-x)({ y }^{ 2 }+{ x }^{ 2 }) }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { (y-x)({ y }^{ 2 }+{ x }^{ 2 }) }{ x } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { { y }^{ 3 }-xy^{ 2 }+yx^{ 2 }-{ x }^{ 3 }-{ y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }$
$=\dfrac { \dfrac { -x(y^{ 2 }-xy+{ x }^{ 2 }) }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =-\dfrac { y^{ 2 }-xy+{ x }^{ 2 } }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =-1$

Hence, $\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }=-1$  

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $\dfrac {x}{y}=\dfrac {3}{4}$ and $\dfrac {x}{2z}=\dfrac {3}{2}$, then $\dfrac {2x+z}{x-2z}+\left (\dfrac {6}{7}+\dfrac {y-x}{y+x}\right )$ will be equivalent to

  1. 8

  2. 9

  3. 11

  4. 12
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\because \dfrac {x}{y}=\dfrac {3}{4}\Rightarrow \dfrac {y-x}{y+x}=\dfrac{4-3}{4+3}=\dfrac {1}{7}$
$\because \dfrac {x}{2z}=\dfrac {3}{2}\Rightarrow 2x=6z$ or, $x=3z$
$\therefore \dfrac {2x+z}{x-2z}+\left (\dfrac {6}{7}+\dfrac {y-x}{y+x}\right )=\dfrac {6z+z}{3z-2z}+\left (\dfrac {6}{7}+\dfrac {1}{7}\right )$
$=\dfrac {7z}{z}+\dfrac {7}{7}=7+1=8$.

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

Here 'x' in the following is : $\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=b$

  1. $\dfrac{2ab}{(b^2+1)}$
  2. $\dfrac{2ab}{a+b}$
  3. $\dfrac{a+b}{2ab}$
  4. $\dfrac{b^2+1}{2ab}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-{\sqrt{a-x}}}=\dfrac{b}{1}$

$\Rightarrow \dfrac{\sqrt{a+x}}{\sqrt{a-x}}=\dfrac{b+1}{b-1}$

$\Rightarrow \dfrac{a+x}{a-x}=\dfrac{(b+1)^2}{(b-1)^2}$

$\Rightarrow \dfrac{(a+x)+(a-x)}{(a+x)-(a-x)}=\dfrac{(b+1)^2+(b-1)^2}{(b+1)^2-(b-1)^2}$

$\Rightarrow \dfrac{2a}{2x}=\dfrac{2(b^2+1)}{4b}$

$\Rightarrow x=\dfrac{2ab}{b^2+1}$

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

 $\dfrac{a+be^y}{a-be^y} = \dfrac{b+ce^y}{b-ce^y}  =  \dfrac{c+de^y}{c-de^y}$, then $a,b,c,d$  are  in

  1. $A.P.$
  2. $G.P.$
  3. $H.P.$
  4. $A.G.P.$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the property of componendo and dividendo on the given equation (a+be^y)/(a-be^y) = (b+ce^y)/(b-ce^y), we get a/(be^y) = b/(ce^y). This simplifies to a/b = b/c, which implies b^2 = ac, meaning a, b, and c are in G.P. Extending this to the third term confirms a, b, c, and d are in G.P.

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

If $y^2 = ax^2 +bx+c$, then $y^2 \dfrac{d^2y}{dx^2}$ is

  1. a constant function

  2. a function of x only

  3. a function of y only

  4. a function of both x and y
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Differentiating y^2 = ax^2 + bx + c gives 2y*y' = 2ax + b. Differentiating again gives 2(y')^2 + 2y*y'' = 2a. Substituting y' = (2ax+b)/(2y) into the equation allows one to solve for y^2*y''. The result is a constant.

Multiple choice maths numbers in indian and international systems indian system of numeration numbers to 1 million formation of large numbers

If y is an implicit function of x defined by ${ x }^{ 2x }-{ 2x }^{ x }coty-1=0.$ Then, $y' (1)$ is equal to

  1. $-1$
  2. $1$
  3. $\log 2$
  4. $-\log 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ x }^{ 2x }-{ 2x }^{ x }coty-1=0.$ ...............[1]


At $x=1$; we have


$1-\cot y-1=0$

$\implies y=\dfrac{\pi}{2}$

Differentiating w.r.t $x$, we get:

$2x^{2x}(1+\ln x)-2[x^x(-cosec^2y\dfrac{dy}{dx}+\cot yx^x(1+\ln x))]=0$

At $P(1,\dfrac{\pi}{2})$, we have

$2(1+\ln 1)-2[1(-1)\dfrac{dy}{dx}| _P+0]=0$

$\implies 2+2\dfrac{dy}{dx}| _P=0$

$\implies \dfrac{dy}{dx}| _P=-1$

Hence, $y'(1)=-1$

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $\log \sqrt{x^2+y^2}=\tan^{-1}\left(\dfrac{y}{x}\right)$ , then $\dfrac{dy}{dx}$ is:

  1. $1$
  2. $2$
  3. $\dfrac{2x}{\sqrt{x^2+y^2}}$
  4. $\dfrac{x+y}{x-y}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given, 
$\log\sqrt{x^2+y^2}=\tan ^{-1}\left(\dfrac{y}{x}\right)$
Now, differentiating both sides w.r.to $x$ we get,
or, $\dfrac{1}{2}\dfrac{2x+2y\dfrac{dy}{dx}}{x^2+y^2}=\dfrac{x^2}{x^2+y^2}.\left(-\dfrac{y}{x^2}+\dfrac{\dfrac{dy}{dx}}{x}\right)$
or, $x+y\dfrac{dy}{dx}=-y+x\dfrac{dy}{dx}$
or, $(x-y)\dfrac{dy}{dx}=(x+y)$
or, $\dfrac{dy}{dx}=\dfrac{x+y}{x-y}$
Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $y = \left| {\cos x} \right| + \left| {\sin x} \right|$ , then ${\dfrac{dy} {dx}}$ at $x = {\dfrac {2\pi } 3}$ is

  1. ${1 \over 2}\left( {\sqrt 3 + 1} \right)$
  2. $2\left( {\sqrt 3 - 1} \right)$
  3. ${1 \over 2}\left( {\sqrt 3 - 1} \right)$

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

$y=|\cos x|+|\sin x|$


at  $x=\dfrac{2\pi}{3},$

$y=-\cos x+\sin x$

$now, \dfrac{d{y}}{d{x}}=\sin x+\cos x$

$\dfrac{d{y}}{d{x}}=\sin\dfrac{2\pi}{3} +\cos \dfrac{2\pi}{3}$

$ \dfrac{d{y}}{d{x}}=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}$

$ \dfrac{d{y}}{d{x}}=\dfrac{1}{2}(\sqrt{3}+1)$

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ e }^{ t }$ where $t=\sin ^{ -1 }{ \left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right)  } $ then $\cfrac { dy }{ dx } $ is equal to

  1. $\cfrac { x-y }{ x+y } $
  2. $\cfrac { x+y }{ x-y } $
  3. $\cfrac { y-x }{ y+x } $
  4. $\cfrac { x-y }{ 2x+y } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$t=\sin ^{ -1 }{ \left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right)  } $, differentiate on both sides.
$\cfrac { dt }{ dx } =\cfrac { 1 }{ \sqrt { 1-{ \left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right)  }^{ 2 } }  } \cfrac { d\left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right)  }{ dx } \quad \left( \because \cfrac { d\left( \sin ^{ -1 }{ x }  \right)  }{ dx } =\cfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  }  \right) $
$=\left( \cfrac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }{ x }  \right) \left[ \cfrac { \left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  \right) \cfrac { dy }{ dx } -\left( \cfrac { 2x+xy\cfrac { dy }{ dx }  }{ x\sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right) y }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right)  }  \right] $
$=\left( \cfrac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }{ x }  \right) \left[ \cfrac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \cfrac { dy }{ dx } -\left( xy+{ y }^{ 2 }\cfrac { dy }{ dx }  \right)  }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  }  \right] $
$\therefore \cfrac { dt }{ dx } =\cfrac { { x }^{ 2 }\cfrac { dy }{ dx } -xy }{ x\left( { x }^{ 2 }+{ y }^{ 2 } \right)  } =\cfrac { x\cfrac { dy }{ dx } -y }{ { x }^{ 2 }+{ y }^{ 2 } } $
Given that $\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ e }^{ t }$
${ x }^{ 2 }+{ y }^{ 2 }={ e }^{ 2t }$ differentiate on both sides
$2xx+2y\cfrac { dy }{ dx } ={ e }^{ 2t }\left( 2 \right) \cfrac { dt }{ dx } $
$x+y\cfrac { dy }{ dx } =\left( \quad { x }^{ 2 }+{ y }^{ 2 } \right) \left( \cfrac { x\cfrac { dy }{ dx } -y }{ { x }^{ 2 }+{ y }^{ 2 } }  \right) $
$x+y=(x-y)\cfrac { dy }{ dx } $
$\therefore \cfrac { dy }{ dx } =\cfrac { x+y }{ x-y } $
Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $y = \dfrac { 1 } { 1 + x ^ { n - m } + x ^ { p - m } } + \dfrac { 1 } { 1 + x ^ { m - n } + x ^ { p - n } } + \dfrac { 1 } { 1 + x ^ { m - p } + x ^ { n - p } }$ then $\dfrac { d y } { d x }$ at $x = e ^ { m ^ { n p } }$ is equal to

  1. $e ^ { m n p }$
  2. $e ^ { m n / p }$
  3. $e ^ { n p / m }$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\\y=(\dfrac{1}{1+(\dfrac{x^n}{x^m})+(\dfrac{x^p}{x^m})})+(\dfrac{1}{1+(\dfrac{x^m}{x^n})+(\dfrac{x^p}{x^n})})+(\dfrac{1}{1+(\dfrac{x^m}{x^p})+(\dfrac{x^n}{x^p})})$

$\\=(\dfrac{x^m}{x^m+x^n+x^p})+(\dfrac{x^n}{x^n+x^m+x^p})+(\dfrac{x^p}{x^p+x^m+x^n})$

$\\=(\dfrac{x^m+x^n+x^p}{x^m+x^n+x^p})$

$\\=1$
$\\\therefore\>(\dfrac{dy}{dx})=0$
Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If f'$\left( x \right) =\sqrt { { 2x }^{ 2 }-1 } $ and y=f$\left( { x }^{ 2 } \right) $ then $\dfrac { dy }{ dx } $ at x=1 is

  1. 2

  2. 1

  3. -2

  4. none of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

$f\left( x \right)=\sqrt{2{{x}^{2}}-1}$

And

$ y=f\left( {{x}^{2}} \right) $

$ y={{\left( \sqrt{2{{x}^{2}}-1} \right)}^{2}} $

$ y=2{{x}^{2}}-1 $


On differentiating and we get,

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

$ \dfrac{dy}{dx}=4x-0 $

$ \dfrac{dy}{dx}=4x $

At point $\left( x=1 \right)$

So,

$ \dfrac{dy}{dx}=4x=4\left( 1 \right) $

$ \dfrac{dy}{dx}=4 $

Hence, this is the answer.

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $y = \dfrac {1}{1 + x^{n - m} + x^{p - m}} + \dfrac {1}{1 + x^{m - n} + x^{p - n}} + \dfrac {1}{1 + x^{m - p} +x^{n - p}}$ then $\dfrac {dy}{dx}$ at $e^{m^{n^{p}}}$ is equal to

  1. $e^{mnp}$
  2. $e^{mn/p}$
  3. $e^{np/m}$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\quad \quad y=\cfrac { 1 }{ 1+{ x }^{ n-m }+{ x }^{ p-m } } +\cfrac { 1 }{ 1+{ x }^{ m-n }+{ x }^{ p-n } } +\cfrac { 1 }{ 1+{ x }^{ m-p }+{ x }^{ n-p } } \\ \Rightarrow y=\cfrac { 1 }{ 1+\cfrac { { x }^{ n } }{ { x }^{ m } } +\cfrac { { x }^{ p } }{ { x }^{ m } }  } +\cfrac { 1 }{ 1+\cfrac { { x }^{ m } }{ { x }^{ n } } +\cfrac { { x }^{ p } }{ { x }^{ n } }  } +\cfrac { 1 }{ 1+\cfrac { { x }^{ m } }{ { x }^{ p } } +\cfrac { { x }^{ n } }{ { x }^{ p } }  } $

$\Rightarrow y=\cfrac { { x }^{ m } }{ { x }^{ m }+{ x }^{ n }+{ x }^{ p } } +\cfrac { { x }^{ n } }{ { x }^{ m }+{ x }^{ n }+{ x }^{ p } } +\cfrac { { x }^{ p } }{ { x }^{ m }+{ x }^{ n }+{ x }^{ p } } $

$\quad \quad \quad =\cfrac { 1 }{ \left( { x }^{ m }+{ x }^{ n }+{ x }^{ p } \right)  } \left( { x }^{ m }+{ x }^{ n }+{ x }^{ p } \right) $

$\Rightarrow y=1\\ \Rightarrow \cfrac { dy }{ dx } =0\quad $ at all $x$

$\therefore \dfrac { dy }{ dx } $ at ${ e }^{  m ^ { n ^{ p } } }=0$

D answer

 

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

if $y = 500{e^{7x}} + 600{e^{-7x}}$ . Then $\dfrac{{d^2y}}{dx^2} = 49y$

  1. True

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

$y=500e^{7x}+600e^{-7x}$


Differentiationg w.r.t. $x$

$\Rightarrow$  $\dfrac{dy}{dx}=500\dfrac{d(e^{7x})}{dx}+600\dfrac{d(e^{-7x})}{dx}$

$\Rightarrow$  $500\times e^{7x}\times \dfrac{d(7x)}{dx}+600\times e^{-7x}\times \dfrac{d(-7x)}{dx}$

$\Rightarrow$  $\dfrac{dy}{dx}=500\times e^{7x}\times 7+600\times e^{-7x}\times (-7)$

$\Rightarrow$  $\dfrac{dy}{dx}=500\times 7\times e^{7x}-600\times 7\times e^{-7x}$

Again differentiating w.r.t. $x$,

$\Rightarrow$  $\dfrac{d^2y}{dx^2}=500\times 7\times \dfrac{d(e^{7x})}{dx}-600\times 7\times \dfrac{d(e^{-7x})}{dx}$

$\Rightarrow$  $\dfrac{d^2y}{dx^2}=500\times 7\times e^{7x}\times 7-600\times 7\times (-7)\times e^{-7x}$

$\Rightarrow$  $\dfrac{d^2y}{dx^2}=500\times 7\times 7e^{7x}+600\times 7\times 7\times e^{-7x}$

$\Rightarrow$  $\dfrac{d^2y}{dx^2}=7\times 7(500e^{7x}+600e^{-7x})$

$\Rightarrow$  $\dfrac{d^2y}{dx^x}=49y$