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 general knowledge math & puzzles

d/dx(1/u^n) = (^ is Power)

  1. -(v/u^n+1)dy/dx

  2. (n/n^u+1)dv/dx

  3. -(n/u^n+1)du/dx

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the power rule for differentiation, d/dx(u^-n) = -n * u^(-n-1) * du/dx. This simplifies to -(n/u^(n+1)) * du/dx. The other options either use the wrong variable (v), incorrect power notation, or omit the chain rule component (du/dx).

Multiple choice maths indices negative indices law of indices laws of indices

$(x+y)^{-1}(x^{-1}+y^{-1})$ is the reciprocal of

  1. $x+y$
  2. $xy$
  3. $\displaystyle\frac{1}{xy}$
  4. $\displaystyle\frac{x}{y}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle(x+y)^{-1}(x^{-1}+y^{-1})=\frac{1}{x+y}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x+y}\times\frac{x+y}{xy}=\frac{1}{xy}$

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

If $y=\dfrac{1+x^2+x^4}{1+x+x^2}$ and $\dfrac{dy}{dx}=ax+b$, then the values of $a$ and $b$ are,

  1. $a=2,b=1$
  2. $a=-2,b=1$
  3. $a=2,b=-1$
  4. $a=-2,b=-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\displaystyle y=\frac{1+x^{2}+x^{4}}{1+x+x^{2}}$

$\displaystyle \frac{dy}{dx}= \frac{(2x+4x^{3})(1+x+x^{2})-(1+2x)(1+x^{2}+x^{4})}{(1+x^{2}+x^{4}+2x+2x^{2}+2x^{3})}$

$\displaystyle \frac{dy}{dx}=\frac{2x+2x^{2}+2x^{3}+4x^{3}+4x^{4}+4x^{5}-1-x^{2}-x^{4}-2x-2x^{3}-2x^{5}}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle \frac{dy}{dx}=\frac{2x^{5}+3x^{4}+4x^{3}+x^{2}-1}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle (\frac{dy}{dx})=\frac{(2x^{5}+4x^{4}+6x^{3}+4x^{2}+2x)-(x^{4}+2x^{3}+3x^{2}+2x+1)}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle (\frac{dy}{dx})=\frac{2x[x^{4}+2x^{3}+3x^{2}+2x+1]}{(x^{4}+2x^{3}+3x^{2}+2x+1)}-1$

$\displaystyle (\frac{dy}{dx})=2x-1= ax+b$

$\Rightarrow a=2$ & $b=-1$
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

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 ${ x }^{ 2 }.{ e }^{ y }+2x{ ye }^{ x }+13=0$, then $\dfrac { dy }{ dx }$ is

  1. $\dfrac { -2x{ e }^{ y-x }-2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$
  2. $\dfrac { 2x{ e }^{ x-y }+2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$
  3. $\dfrac { -2x{ e }^{ x-y }+2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Differentiating w.r.t. $x$

$2xe^y+x^2e^y\dfrac{dy}{dx}+2[\dfrac{d}{dx}(xy)e^x+xye^x]=0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2[\dfrac{d}{{dx}}(xy){e^x} + xy{e^x}] = 0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2{e^x}[y + x\dfrac{{dy}}{{dx}} + xy] = 0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2y{e^x} + 2x{e^x}\dfrac{{dy}}{{dx}} + 2xy2{e^x} = 0$

$\left( {{x^2}{e^y} + 2x{e^x}} \right)\dfrac{{dy}}{{dx}} =  - \left( {2x{e^y} + 2y{e^x} + 2xy{e^x}} \right)$

$\dfrac{{dy}}{{dx}} =  - \dfrac{{\left( {2x{e^y} + 2y{e^x} + 2xy{e^x}} \right)}}{{{x^2}{e^y} + 2x{e^x}}}$

$\dfrac{{dy}}{{dx}} =  - \dfrac{{\left( {2x{e^{y - x}} + 2y + 2xy} \right)}}{{{x^2}{e^{y - x}} + 2x}}$

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


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

Find: $\dfrac{d}{{\text dx}}\left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right)  \,\,\,$

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

Let $ y = \dfrac{1- \cos x }{\sin x}$


Formula: $\dfrac{d \left (\dfrac{u} {v}\right)}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}$

differentiate on both sides w.r.t $x$


$\dfrac{dy}{dx} = \dfrac{(\sin x)\sin x- (1- \cos x)\cos x}{(\sin x)^{2}}$

$= \dfrac{\sin ^{2}x -cos x + \cos ^{2}x}{(\sin x)^{2}}$

$= \dfrac{1-\cos x}{(\sin x)^{2}}$

$=  \dfrac{2 \sin^{2}\dfrac{x}{2}}{4\sin^{2}\dfrac{x}{2}\cos^{2}\dfrac{x}{2}}$

$= \dfrac{1}{2}\sec^{2}\left(\dfrac{x}{2}\right)$

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

If $xe^{xy}-y=\sin x$, then $\dfrac {dy}{dx}$ at $x=0$ is

  1. $0$
  2. $1$
  3. $-1$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$x _e^ {xy}-y=\sin x$
Differentiating both sides with respect to $x$,
$x\dfrac {d}{dx} (e^{xy}) -\dfrac {d}{dy}+e^{xy}\dfrac {d}{dx}(x)=\dfrac {d}{dx} (\sin x)$
or, $xe^{xy}\dfrac {d}{dx}(xy)-\dfrac {dy}{dx}+e^{xy}=\cos x$
or, $xe^{xy}y+xe^{xy}\dfrac {dy}{dx}-\dfrac {dy}{dx} +e^{xy} +e^{xy}=\cos x$
or, $\dfrac {dy}{dx} (x e^{xy}-1)=\cos x e^{xy} -xye^{xy}$
$\therefore \ \left. \dfrac {dy}{dx}\right] _{x=0}=\dfrac {\cos x-e^{xy} -xye^{xy}}{xe^{xy}-1}=\dfrac {1-0}{0-1}=-1$