Mathematics

Differentiation and Derivatives

60 Questions

Differentiation involves finding the rate of change of functions using established mathematical rules. Topics include the chain rule, quotient rule, and calculating partial derivatives for multivariable equations. Mastery of these mathematical calculations is heavily tested in undergraduate entrance and competitive exams.

Chain rule derivativesQuotient rule applicationPartial derivatives calculationDirectional derivative vectorsLogarithmic function derivatives

Differentiation and Derivatives Questions

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

Differentiate $\tan^{-1} \sqrt{\dfrac{1+\cos x}{1- \cos x}}$

  1. $\dfrac{-1}{2}$
  2. $\dfrac{1}{4}$
  3. $\dfrac{1}{8}$

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

Let $y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 + \cos x}}{{1 - \cos x}}} $

$ = {\tan ^{ - 1}}\sqrt {\dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2{{\sin }^2}\dfrac{x}{2}}}} $
$ = {\tan ^{ - 1}}\cot \left( {\dfrac{x}{2}} \right)$
$ = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right)} \right]$
$ = \dfrac{\pi }{2} - \dfrac{x}{2}$
$ \Rightarrow \dfrac{{dy}}{{dx}} =  - \dfrac{1}{2}$

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

Differential coefficient of $\log\ \sin x$ is :

  1. $\cos x$
  2. $\tan x$
  3. $\text{cosec} \,x$
  4. $\cot x$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

$y=\log \sin x$

On differentiating w.r.t $x$, we get
$\dfrac{dy}{dx}=\dfrac{d(\log \sin x)}{dx}$
$\dfrac{dy}{dx}=\dfrac{1}{\sin x}\times \cos x$
$\dfrac{dy}{dx}=\dfrac{\cos x}{\sin x}$
$\dfrac{dy}{dx}=\cot x$

Hence, this is the answer.

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

Derivative of $(\sin x)^x + \sin^{-1} \sqrt{x}$ with respect to $x$ is

  1. $(x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$
  2. $(x \cot x + \log \sin x) + \dfrac{1}{\sqrt{x - x^2}}$
  3. $(\sin x)^x (x \cot x + \log \,x) + \dfrac{1}{\sqrt{x - x^2}}$
  4. $(\sin x)^x (x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $y=(\sin x)^x$

$\Rightarrow \log y=x \log (\sin x)$
Now differentiating both sides with respect to $x$ 
$\dfrac{1}{y}\dfrac{dy}{dx}=x\cot x+\log \sin x$
or, $\dfrac{dy}{dx}=(\sin x )^x{x\cot x+\log \sin x}$........(1).
Again let $z=(\sin x)^x+\sin^{-1}\sqrt{x}$
Now differentiating both sides with respect to $x$.
$\dfrac{dz}{dx}=\dfrac{dy}{dx}+\dfrac{1}{\sqrt{1-x}}.\dfrac{1}{2\sqrt{x}}$
$\dfrac{dz}{dx}=(\sin x)^x{x\cot x+\log \sin x}+\dfrac{1}{2\sqrt{x-x^2}}$ [Using (1)]

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

Derivative of ${ \log { x }  }^{ \cos { x }  }$ with respect to $x$ is

  1. ${ \log { x } }^{ \cos { x } }\left[ \cfrac { \cos { x } }{ x\log { x } } -\sin { x } \log { \left( \log { x } \right) } \right] \quad $
  2. ${ \log { x } }^{ \cos { x } }\left[ \cfrac { \cos { x } }{ x\log { x } } -\cos { x } \log { \left( \log { x } \right) } \right] \quad $
  3. ${ \log { x } }^{ \sin { x } }\left[ \cfrac { \sin { x } }{ x\log { x } } -\sin { x } \log { \left( \log { x } \right) } \right] \quad $

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $y=\log x^{\cos x}$

$\Rightarrow \ y=(\cos x)(\log x)$

$\therefore \dfrac {dy}{dx}=(\cos x) \left (\dfrac {1}{x}\right)+(-\sin x)(\log x)$

$=\dfrac {\cos x}{x}-\sin x\log x$
Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $y = \log \left( \frac { 1 + x } { 1 - x } \right) ^ { 1 / 4 } - \frac { 1 } { 2 } \tan ^ { - 1 } x ,$ then $\frac { d y } { d x } =$

  1. $\frac { x ^ { 2 } } { 1 - x ^ { 4 } }$
  2. $\frac {2 x ^ { 2 } } { 1 - x ^ { 4 } }$
  3. $\frac { x ^ { 2 } } { 2 \left( 1 - x ^ { 4 } \right) }$

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

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


$\dfrac{1-x}{4(1+x)}\times\dfrac{2}{(1-x)^2}$ $-\dfrac{1}{2(x^2+1)}$  $=\dfrac{1}{2(1-x^2)}-$ $\dfrac{1}{2(x^2+1)}$ =$\dfrac{x^2}{1-x^4}$

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

The value of $\displaystyle \frac{d}{dx} (|x-1|+ |x-5|) $ at x = 3 is

  1. -2

  2. 0

  3. 2

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

$\dfrac{d}{dx} |x| = \dfrac{|x|}{x}$


Applying the above formula ,

$\dfrac{d}{dx} ( |x-1|+|x-5|) = \dfrac{d}{dx}|x-1|+\dfrac{d}{dx}|x-5|= \dfrac{|x-1|}{x-1}+\dfrac{|x-5|}{x-5}$

Substituting $x=3$,

$\dfrac{|3-1|}{3-1}+\dfrac{|3-5|}{3-5} = \dfrac{|2|}{2} +\dfrac{|-2|}{-2} = \dfrac{2}{2} +\dfrac{2}{-2} =1-1 =0$

Multiple choice

What is the derivative of (sin(x)) with respect to (x)?

  1. \(cos(x)\)
  2. \(sec(x)\)
  3. \(-sin(x)\)
  4. \(cosec(x)\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The derivative of (sin(x)) with respect to (x) is (cos(x)).

Multiple choice

What is the exterior derivative of a differential form?

  1. The gradient of the differential form

  2. The divergence of the differential form

  3. The curl of the differential form

  4. The Laplacian of the differential form
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

The exterior derivative of a differential form is a new differential form that measures the boundary of the form.

Multiple choice

What is the purpose of using numerical differentiation in applications?

  1. Estimating the slope of a function at a given point

  2. Finding the roots of polynomials

  3. Solving differential equations

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

Numerical differentiation is used to approximate the derivative of a function at a given point using numerical methods, which is useful when the analytical derivative is difficult or impossible to obtain.

Multiple choice

What is the derivative of the function (f(x) = x^3 + 2x^2 - 3x + 4)?

  1. 3x^2 + 4x - 3

  2. 3x^2 + 4x + 3

  3. 3x^2 - 4x - 3

  4. 3x^2 - 4x + 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The derivative of a function (f(x)) is given by the limit (\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}). Applying this limit to the function (f(x) = x^3 + 2x^2 - 3x + 4), we get: (\lim_{h \to 0} \frac{(x+h)^3 + 2(x+h)^2 - 3(x+h) + 4 - (x^3 + 2x^2 - 3x + 4)}{h}) = (\lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 + 2x^2 + 4xh + 2h^2 - 3x - 3h + 4 - x^3 - 2x^2 + 3x - 4}{h}) = (\lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 + 4xh + 2h^2 - 3h}{h}) = (\lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 + 4x + 2h - 3)}{h}) = (\lim_{h \to 0} (3x^2 + 3xh + h^2 + 4x + 2h - 3)) = (3x^2 + 4x - 3). Therefore, the derivative of the function (f(x) = x^3 + 2x^2 - 3x + 4) is (3x^2 + 4x - 3).

Multiple choice

What is the derivative of the function f(x) = x^3 - 2x^2 + 3x - 4?

  1. 3x^2 - 4x + 3

  2. 3x^2 - 2x + 3

  3. 3x^2 - 4x + 1

  4. 3x^2 - 2x + 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The derivative of a function f(x) is given by the formula f'(x) = lim_(h->0) [f(x+h) - f(x)] / h. Substituting f(x) = x^3 - 2x^2 + 3x - 4, we get f'(x) = lim_(h->0) [(x+h)^3 - 2(x+h)^2 + 3(x+h) - 4 - (x^3 - 2x^2 + 3x - 4)] / h = lim_(h->0) [x^3 + 3x^2h + 3xh^2 + h^3 - 2x^2 - 4xh - 2h^2 + 3x + 3h - 4 - x^3 + 2x^2 - 3x + 4] / h = lim_(h->0) [3x^2h + 3xh^2 + h^3 - 4xh - 2h^2 + 3h] / h = lim_(h->0) [3x^2 + 3xh + h^2 - 4x - 2h + 3] = 3x^2 - 4x + 3.

Multiple choice

What is the derivative of the function f(x) = x^3 - 2x^2 + 3x - 4?

  1. 3x^2 - 4x + 3

  2. 3x^2 - 4x + 1

  3. 3x^2 - 2x + 3

  4. 3x^2 - 2x + 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The derivative of a function f(x) is given by the formula f'(x) = lim_(h->0) [f(x+h) - f(x)] / h. Substituting the values of f(x), we get f'(x) = lim_(h->0) [(x+h)^3 - 2(x+h)^2 + 3(x+h) - 4 - (x^3 - 2x^2 + 3x - 4)] / h. Simplifying this equation, we get f'(x) = 3x^2 - 4x + 3.

Multiple choice

What is the name of the theorem that states that the derivative of an inverse function is equal to the reciprocal of the derivative of the original function?

  1. Inverse function theorem

  2. Chain rule

  3. Mean value theorem

  4. Cauchy's mean value theorem
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The inverse function theorem states that the derivative of an inverse function is equal to the reciprocal of the derivative of the original function.