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Preview Of Differential Calculus

Learn preview of differential calculus and meaning of derivative of functions. Practice to double derivative, and standard trigonometric functions, chain rule and product rule derivative equation.

Meaning of Derivative 

  • Let \(y\) be a function of \(x\) i.e., \(y=f(x)\) and \(\Delta y\) be the change in \(y\) corresponding to a small change \((\Delta x)\) in \(x\).
  • Then, \(\dfrac{\Delta y}{\Delta x}\) represents the average rate of change of \(y\) with respect to \(x\) as \(x\) changes  from \(x\) to \(x + \Delta x\).
  • As \(\Delta x \to 0\), the limiting  value of this average rate of change of \(y\) with respect to \(x\) in the interval  \([x, x + \Delta x]\) becomes the instantaneous rate of change of \(y\) with respect to \(x\) i.e., \(\left (\dfrac{dy}{dx}\right)\)
  • Thus, \(\lim \limits_{\Delta x \to 0} \dfrac{\Delta y}{\Delta x}\) = Instantaneous rate of change of \(y\) with respect to \(x= \dfrac{dy}{dx}\)

             \(\Rightarrow\) \( \dfrac{dy}{dx}\) = Instantaneous rate of change of \(y\) with respect to \(x\)

                    \(\left[\because\, \lim \limits_{\Delta x \to 0} \dfrac{\Delta y}{\Delta x}= \dfrac{dy}{dx}\right]\)

  • Hence, to determine the instantaneous rate of change of one quantity with respect to another quantity, calculate its derivative.
  • Suppose the position \((x)\) of a particle is given as the function of time \((t)\) while it is moving along a straight line.

            Then, average velocity = \(\dfrac{\Delta x}{\Delta t}\) 

and, instantaneous velocity = \(\dfrac{d x}{d t}\)

  • To determine the instantaneous acceleration, take the derivative of instantaneous velocity \(\left (\dfrac{dx}{dt}\right)\) i.e., \(a_{inst} = \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)\)

            \(a_{inst} = \dfrac{d}{dt}\left(v_{inst}\right)\) 

            \(a_{inst} = \dfrac{d \,v_{inst}}{dt}\)              

 

          

Illustration Questions

Which of the following statement is correct? 

A Velocity is a derivative of position with respect to time

B Velocity is a derivative of acceleration with respect to time

C Velocity is a derivative of position with respect to velocity

D Velocity is a derivative of acceleration with respect to position

×

Since, velocity is a derivative of position with respect to time.

Thus, option (A) is correct.

Which of the following statement is correct? 

A

Velocity is a derivative of position with respect to time

.

B

Velocity is a derivative of acceleration with respect to time

C

Velocity is a derivative of position with respect to velocity

D

Velocity is a derivative of acceleration with respect to position

Option A is Correct

Derivative of xn

  • Let \(y\) be a function of \(x\), as given

           \(y= f(x) = x^n\)

where 'n' is a real number

            Then, the derivative of the function \(y= f(x)\) will be  

           \(\dfrac{dy}{dx} = n x^{n-1}\)

Illustration Questions

 A particle is moving along a straight line. Its position, as a function of time, is given as  \(x = t^3 + 3t\). Calculate its velocity.

A \(3 t ^2 + 3\)

B \(9 t + 6\)

C \(10 t +5\)

D \( t + 3\)

×

\(v = \dfrac{d}{dt} (x)\)

\(x= t^3 + 3t\)

\(v = \dfrac{d}{dt} (t^3 + 3t)\)

  \(= \dfrac{d}{dt} (t^3) + 3\dfrac{d}{dt} (t)\)

  \(= 3t^2 + 3\)

 A particle is moving along a straight line. Its position, as a function of time, is given as  \(x = t^3 + 3t\). Calculate its velocity.

A

\(3 t ^2 + 3\)

.

B

\(9 t + 6\)

C

\(10 t +5\)

D

\( t + 3\)

Option A is Correct

Derivative of some Standard Trigonometric Functions 

  • \(\dfrac{d}{dx} (sin \, x) = cos \,x\)
  • \(\dfrac{d}{dx} (cos\, x) = -sin \,x\)
  • \(\dfrac{d}{dx} (tan\, x) = sec^2 \,x\)
  • \(\dfrac{d}{dx} (cot\, x) = -cosec^2\,x\)
  • \(\dfrac{d}{dx} (sec\, x) = sec \,x \, tan \,x\)
  • \(\dfrac{d}{dx} (cosec\, x) = -cosec \,x \, cot \,x\)

Illustration Questions

The derivative of \(y = 2\, sin \,x + 5\, tan \,x \) is

A \(2 \, sin \,x + 5\; sec^2 \,x\)

B \(3 \, cos \,x + 6\, sin \,x\)

C \(2 \, cos \,x + 5 \;sec^2 \,x\)

D \(5 \, sec \,x \)

×

\(y= 2 \, sin \,x + 5 \, tan \,x\)

\(\dfrac{dy}{dx} = \dfrac{d}{dx} (2\,sin\,x) + \dfrac{d}{dx} (5\,tan\,x)\)

\(\dfrac{dy}{dx} = 2\,cos\,x + 5\,sec^2\,x\)

The derivative of \(y = 2\, sin \,x + 5\, tan \,x \) is

A

\(2 \, sin \,x + 5\; sec^2 \,x\)

.

B

\(3 \, cos \,x + 6\, sin \,x\)

C

\(2 \, cos \,x + 5 \;sec^2 \,x\)

D

\(5 \, sec \,x \)

Option C is Correct

Derivative of Exponential and Logarithmic Function 

  • \(\dfrac{d}{dx}(e^x) = e^x\)
  • \(\dfrac{d}{dx}(\ell n\,x) = \dfrac{1}{x}\)

Illustration Questions

The derivative of \(y= 2\,e ^x + \ell n \,x \) is

A \(2 \,e^x + x\)

B \(3 \,e^x + 2\,x\)

C \(3 \,e^x + 6\,x\)

D \(2 \,e^x + \dfrac{1}{x}\)

×

\(y = 2\, e^x + \ell n \,x\)

\(\dfrac{dy}{dx} = 2\dfrac{d(e^x)}{dx} + \dfrac{d(\ell n\,x)}{dx}\)

\(\dfrac{dy}{dx} = 2 \, e ^x + \dfrac{1}{x} \)

The derivative of \(y= 2\,e ^x + \ell n \,x \) is

A

\(2 \,e^x + x\)

.

B

\(3 \,e^x + 2\,x\)

C

\(3 \,e^x + 6\,x\)

D

\(2 \,e^x + \dfrac{1}{x}\)

Option D is Correct

Illustration Questions

The derivative of \(y= x^2 + 4 \, sin \,x + 5 \,e^x\) is

A \(2\,e^x + 6\,x\)

B \(e^x + 3\,cos\,x\)

C \(2\,x + 4\,cos\,x + 5 \,e^x\)

D \(x + e^x\)

×

\(y= x^2 + 4 \, sin \,x + 5 \,e^x\)

\(\dfrac{dy}{dx} = \dfrac{d}{dx} (x^2) + 4\dfrac{d}{dx} (sin\,x) +5\dfrac{d}{dx} (e^x)\)

\(\dfrac{dy}{dx} = 2\,x + 4\,cos\,x +5\,e^x\)

The derivative of \(y= x^2 + 4 \, sin \,x + 5 \,e^x\) is

A

\(2\,e^x + 6\,x\)

.

B

\(e^x + 3\,cos\,x\)

C

\(2\,x + 4\,cos\,x + 5 \,e^x\)

D

\(x + e^x\)

Option C is Correct

Chain Rule 

  • Let \(y\) be a function of \(x\) and \(x\) be a function of t.

          Then, by chain rule,

                \(\dfrac{dy}{dt} =\dfrac{dy}{dx} . \dfrac{dx}{dt} \)       

Illustration Questions

\(y= e^{3x^2}\), find \(\dfrac{dy}{dx}\).

A \(e^{3x} . x\)

B \(e^{3x^2} (6\,x)\)

C \(2\,x\)

D \(e^x\)

×

Let \(y= e^u \)

where, \(u = 3\,x^2\)

\(\dfrac{dy}{dx} =\dfrac{dy}{du} . \dfrac{du}{dx}\)

       \(= \dfrac{de^u}{du}. \dfrac{d}{dx} (3\,x^2)\)

       \(= e^u . (6\,x)\)

       \(= e^{3x^2} (6\,x)\) 

[\(\text{put}\,\, u=3x^2\)]

\(y= e^{3x^2}\), find \(\dfrac{dy}{dx}\).

A

\(e^{3x} . x\)

.

B

\(e^{3x^2} (6\,x)\)

C

\(2\,x\)

D

\(e^x\)

Option B is Correct

Graphical Meaning of dy/dx where y=f(x)

  • If a tangent to the curve \(y= f(x)\) makes an angle \(\theta\) with \(x - axis\) in the positive direction, then \(\dfrac{dy}{dx}\)= slope of tangent = \(tan \,\theta\)
  • The slope of the tangent to the curve \(y= f(x)\) at the point \((x_1 , y_1)\) is given by \(\left[\dfrac{dy}{dx}\right]_{(x_1,y_1)}\) .

Illustration Questions

Find the slope of the tangent to the curve \(y= 3\,x^2 + 5\,x\) at \(x= 3 \).

A 23

B 25

C 30

D 3

×

\(y= 3\,x^2 + 5\,x\)

\(\dfrac{dy}{dx} = 6\,x +5\)

\(\left[\dfrac{dy}{dx} \right] _{x=3} = 6(3) + 5 \)

                 = \(23\)

\(\therefore\) Slope of tangent at \(x= 3 \) is \(23\).

Find the slope of the tangent to the curve \(y= 3\,x^2 + 5\,x\) at \(x= 3 \).

A

23

.

B

25

C

30

D

3

Option A is Correct

Idea of Maxima and Minima 

  • In the given \(y\) versus \(x \) graph, curve \(y = f(x)\) is shown.
  • The point 'A' represents the highest point of the curve. So, it is the point of local maxima.
  • Point 'B' represents the lowest point of the curve. Hence, it is the point of local minima.
  • The tangents drawn at these two points have zero slope.
  • So, it can be concluded that at maxima and minima,

\(\dfrac{dy}{dx} = 0\)

  • For one - dimensional motion,

           \(v=\dfrac{dx}{dt}\)

         When velocity (\(v\)) becomes zero, displacement \((x)\) is said to be maximum or minimum.

  • Similarly,

            \(a = \dfrac{dv}{dt}\)  

         Here, when acceleration becomes zero, velocity is said to be maximum or minimum.

  • To check whether the values are maximum or minimum, the double derivatives are to be found.
  • Thus, condition of maxima is

             \(\dfrac{dy}{dx} = 0\) 

      and, \(\dfrac{d^2y}{dx^2} < 0\)

  • Condition of minima is

         \(\dfrac{dy}{dx} = 0\)

         and, \(\dfrac{d^2y}{dx^2} > 0\)

Illustration Questions

Given, \(y= x^2 + 3x +1\). Calculate the points of maxima and minima.

A Minima at \(x = \dfrac{-3}{2}\), Maxima does not exist

B Minima at \(x =0\), Maxima  at \(x =3\) 

C Minima at \(x =3\), Maxima  at  \(x =0\)

D Minima does not exist, Maxima at \(x = \dfrac{-3}{2}\)

×

\(y= x^2 + 3x +1\)

\(\dfrac{dy}{dx} = 2x +3\)...................(1)

For maxima / minima

\(\dfrac{dy}{dx} = 0\)

\(2x + 3 = 0\)

\(x = \dfrac{-3}{2}\)

Taking derivative of equation (1),

\( \dfrac{d^2y}{dx^2}= 2 \)

\(\Rightarrow \dfrac{d^2y}{dx^2}>0\)

Thus, value of \(y\) is minimum at \(x = \dfrac{-3}{2}\) and maximum value of \(y\) does not exist.

 

Given, \(y= x^2 + 3x +1\). Calculate the points of maxima and minima.

A

Minima at \(x = \dfrac{-3}{2}\), Maxima does not exist

.

B

Minima at \(x =0\), Maxima  at \(x =3\) 

C

Minima at \(x =3\), Maxima  at  \(x =0\)

D

Minima does not exist, Maxima at \(x = \dfrac{-3}{2}\)

Option A is Correct

Product Rule 

  • If  \(y= uv\)  is a product of two functions \(u\) and \(v\), where \(u= f(x) \) and \(v= g(x)\).
  • Then, derivative of \(y= uv\) is given as

    \(\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\)

Illustration Questions

Find \(\dfrac{dy}{dx}\) of \(y= x\, sin \,x\).

A \(cos \,x + sin \,x\)

B \(3\,cos \,x +2\, sin \,x\)

C \(x\,cos \,x + sin \,x\)

D \( sin \,x\)

×

Let, \(u=x\)

\(v= sin \,x\)

\(y= u \,v\)

\(\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\)

      \(= x\dfrac{d}{dx} (sin\,x)+ (sin\,x)\dfrac{d(x)}{dx} \)

      \(= x\, cos\,x + sin\,x\)

Find \(\dfrac{dy}{dx}\) of \(y= x\, sin \,x\).

A

\(cos \,x + sin \,x\)

.

B

\(3\,cos \,x +2\, sin \,x\)

C

\(x\,cos \,x + sin \,x\)

D

\( sin \,x\)

Option C is Correct

Double Derivative

  • Let \(y\) be the function of \(x\) i.e., \(y= f(x)\).
  • When we differentiate \(y\) with respect to \(x\) two times successively, then this process is called double differentiation.
  •  It is represented as \(\dfrac{d^2y}{dx^2}\)

           \(\dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \left(\dfrac{dy}{dx}\right)\)

  • For example - 

            Instantaneous velocity, \(v= \dfrac{dx}{dt}\) and instantaneous acceleration, \(a= \dfrac{dv}{dt}\)

             \(\Rightarrow a = \dfrac{d}{dt} \left(\dfrac{dx}{dt}\right)\)

             \(\Rightarrow a = \dfrac{d^2x}{dt^2}\)

Illustration Questions

Find \( \dfrac{d^2y}{dx^2}\) of \(y= x^2 + 3x\).

A 1

B 2

C 3

D 4

×

\(y= x^2 + 3x\)

\(\dfrac{dy}{dx} = 2x +3\)

\(\dfrac{d^2y}{dx^2} = 2\)

Find \( \dfrac{d^2y}{dx^2}\) of \(y= x^2 + 3x\).

A

1

.

B

2

C

3

D

4

Option B is Correct

Practice Now