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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., $$a_{inst} = \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)$$

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

$$a_{inst} = \dfrac{d \,v_{inst}}{dt}$$

#### 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

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

#### 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

#### 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,

$$\dfrac{dy}{dt} =\dfrac{dy}{dx} . \dfrac{dx}{dt}$$

#### $$y= e^{3x^2}$$, 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

# Derivative of some Standard Trigonometric Functions

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

#### 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

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

#### 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

# 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 $$\left[\dfrac{dy}{dx}\right]_{(x_1,y_1)}$$ .

#### 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, $$\dfrac{d^2y}{dx^2} > 0$$

#### 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

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

#### 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, $$a= \dfrac{dv}{dt}$$

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

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

#### 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