Informative line

### Derivative

Learn derivatives calculus and practice equation of tangent line & average rate of change calculus formula. Identifying the given limit definition of the derivative and then evaluating the derivative to get the Limit.

# Derivatives

The derivatives of a function $$'f'$$ at a number $$'a'$$ is defined as

$$\lim\limits_{h \to 0}\;\left(\dfrac{f(a+h)-f(a)}{h}\right)$$ ...(1)

and is denoted by $$f'(a)$$ (we are assuming that limit exists).  • This limit is used in tangent & velocity problem, therefore we give it a special name called derivatives.
•  $$f'(a)=\lim\limits_{x \to a}\;\left(\dfrac{f(x)-f(a)}{x-a}\right)$$...(2)   $$\to$$ define $$x-a=h$$ in (1)
• (1) and (2) are different ways of writing the derivatives.
•  Derivatives at  $$x=a=$$ $$f'(a)$$ = Slope of tangent to the graph of  $$'f'$$ at $$(a,\,f(a))$$

#### If $$f(x)=\dfrac{4}{x^2}$$ ,then find $$f'(a)$$.

A $$7\,sin\,a$$

B $$\dfrac{-8}{a^3}$$

C $$\dfrac{200}{a^2}$$

D $$2a$$

×

$$f'(a)=\lim\limits_{h \to 0}\;\left(\dfrac{f(a+h)-f(a)}{h}\right)$$

$$=\lim\limits_{h \to 0}\;\left(\dfrac{\dfrac{4}{(a+h)^2}-\dfrac{4}{a^2}}{h}\right)$$

$$=\lim\limits_{h \to 0}\;\left(\dfrac{4a^2-4(a+h)^2}{ha^2(a+h)^2}\right)$$

$$=\lim\limits_{h \to 0}\;\dfrac{4(-2ah-h^2)}{ha^2(a+h)^2}$$

$$=\lim\limits_{h \to 0}\;\dfrac{4(-2a-h)}{a^2(a+h)^2}$$ ,  (put $$h=0$$)

$$\Rightarrow\,f'(a)=\dfrac{-8}{a^3}$$

### If $$f(x)=\dfrac{4}{x^2}$$ ,then find $$f'(a)$$.

A

$$7\,sin\,a$$

.

B

$$\dfrac{-8}{a^3}$$

C

$$\dfrac{200}{a^2}$$

D

$$2a$$

Option B is Correct

# Identifying the Given Limit as Derivative of a Function at a Certain Value

$$=\lim\limits_{h\to 0}\,\left(\dfrac{f(a+h)-f(a)}{h}\right)=f'(a)$$

• Given any limit in the form of L.H.S, identify the value of $$'a'$$ and the function $$'f'$$.
• e.g. $$=\lim\limits_{h\to 0}\,\left(\dfrac{(2+h)^3-8}{h}\right)$$
• $$=\dfrac{f(a+h)-f(a)}{h}$$

Then,  $$f(a+h)=(2+h)^3$$

$$\Rightarrow\,a=2,\;f(x)=x^3$$

$$\therefore$$ Limit =$$f'(2)=3x^2$$ at $$x=2$$

$$=12$$

#### If $$\lim\limits_{h\to 0}\,\left(\dfrac{\sqrt{1+h}-1}{h}\right)= f'(a)$$  then select the correct function $$'f'$$ and value  of $$'a'$$.

A $$a=5,\;f(x)=x^2$$

B $$a=50,\;f(x)=\dfrac{1}{x}$$

C $$a=1,\;f(x)=x^{{1}/{3}}$$

D $$a=3,\;f(x)=sin\,x$$

×

Given $$f'(a)=\lim\limits_{h\to 0}\,\left(\dfrac{(1+h)^{1/3}-1}{h}\right)$$

We know ,

$$f'(a)=\lim\limits_{h\to 0}\,{\left(\dfrac{f(a+h)-f(a)}{h}\right)}$$

On comparing, we get

$$f(a+h)=(1+h)^{1/3},\;f(a)=1$$

$$a=1,\;f(x)=x^{1/3}$$

### If $$\lim\limits_{h\to 0}\,\left(\dfrac{\sqrt{1+h}-1}{h}\right)= f'(a)$$  then select the correct function $$'f'$$ and value  of $$'a'$$.

A

$$a=5,\;f(x)=x^2$$

.

B

$$a=50,\;f(x)=\dfrac{1}{x}$$

C

$$a=1,\;f(x)=x^{{1}/{3}}$$

D

$$a=3,\;f(x)=sin\,x$$

Option C is Correct

# Rate of Change and Relation with Derivative

Let $$y$$ be a quantity which is a function of $$x$$$$y=f(x)$$. If $$x$$ changes from $$x_1$$ to $$x_2$$ we say.

$$\Delta x =x_2-x_1$$ = change in $$x$$

Corresponding change in $$y=f(x_2)-f(x_1)=\Delta y$$  • The average rate of change of  $$y$$ with respect to $$x$$ is

$$=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}=\dfrac{\Delta y}{\Delta x}$$

• Instantaneous rate of change of $$y$$ with respect to $$x$$ at $$x=x_1$$ is

$$=\lim\limits_{x_2\to x_1}\;\left(\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\right)$$

$$=\lim\limits_{\Delta x\to 0}\;\left(\dfrac{f(x_1+\Delta x)-f(x_1)}{\Delta x}\right)$$  Correlating the above limit with derivative we say, $$f'(a)$$ is the instantaneous rate of change of $$y=f(x)$$ with respect to  $$x$$ when, $$x=a$$.

#### If $$y=\sqrt x$$, then the instantaneous rate of change of  $$y$$ with respect to  $$x$$ at  $$x=4$$ is

A $$\dfrac{7}{8}$$

B $$\dfrac{1}{4}$$

C 19

D –2

×

Instantaneous rate of change at  $$x=4$$ $$=f'(4)$$

where  $$y=f(x)$$

$$f'(x)$$$$=\lim\limits_{h\to 0}\;\dfrac{\sqrt{x+h}-\sqrt x}{h}$$

$$=\lim\limits_{h\to 0}\;\left(\dfrac{(\sqrt{x+h}-\sqrt x)(\sqrt{x+h}+\sqrt x)}{h\left(\sqrt{x+h}+\sqrt x\right)}\right)$$

$$=\lim\limits_{h\to 0}\;\dfrac{h}{h\left(\sqrt{x+h}+\sqrt x\right)}$$

$$=\dfrac{1}{2\sqrt x}$$

$$\Rightarrow$$ $$f'(4)$$ $$=\dfrac{1}{2×2}$$

$$=\dfrac{1}{4}$$

### If $$y=\sqrt x$$, then the instantaneous rate of change of  $$y$$ with respect to  $$x$$ at  $$x=4$$ is

A

$$\dfrac{7}{8}$$

.

B

$$\dfrac{1}{4}$$

C

19

D

–2

Option B is Correct

# Identifying the given Limit as some Derivative to Evaluate

• $$\lim\limits_{x\to 0}\;\left(\dfrac{f(a+h)-f(a)}{x-a}\right)=$$ $$f'(a)$$
• $$\lim\limits_{x\to a}\;\dfrac{f(x)-f(a)}{x-a}=$$ $$f'(a)$$

e.g.  $$\lim\limits_{x\to 2}\;\dfrac{x^{10}-2^{10}}{x-2}=10×2^9$$

because $$f(x)=x^{10},\;a=2$$

$$\therefore$$ $$f'(x)$$ $$=10x^9$$ and $$f'(2)$$ $$=10×2^9$$

#### Find the value of  $$\lim\limits_{x\to 7}\;\left(\dfrac{x^{50}-7^{50}}{x-7}\right)$$

A $$20×8^{21}$$

B $$50×7^{49}$$

C $$2×7^{8}$$

D $$18×5^{19}$$

×

Compare the limit with standard derivative limit.

$$\Rightarrow\,\lim\limits_{x\to 7}\;\left(\dfrac{x^{50}-7^{50}}{x-7}\right)$$

$$=\lim\limits_{x\to a}\;\left(\dfrac{f(x)-f(a)}{x-a}\right)$$

$$a=7,\;f(x)=x^{50}$$

Value of limit = $$f'(a)$$ $$=50x^{49}$$ at  $$x=7$$

$$=50×7^{49}$$

### Find the value of  $$\lim\limits_{x\to 7}\;\left(\dfrac{x^{50}-7^{50}}{x-7}\right)$$

A

$$20×8^{21}$$

.

B

$$50×7^{49}$$

C

$$2×7^{8}$$

D

$$18×5^{19}$$

Option B is Correct

# Equation of Tangent in Terms of the Derivatives

The equation of tangent line to  $$y=f(x)$$  at any point $$P(a,\,f(a))$$ on it is a line passing through $$P(a,\,f(a))$$ and whose slope is $$f'(a)$$.    #### Find the equation of tangent line to the graph of  $$y=f(x)$$  at  $$x=7$$  if $$f(7)=-5$$ and $$f'(7)=4$$.

A $$4x-y-33=0$$

B $$x+2y-1=0$$

C $$x=7$$

D $$y=-8$$

×

Equation of tangent $$\to y-f(a)=f'(a)(x-a)$$

$$\Rightarrow\,y-f(7)=f'(7)(x-(7))$$

$$\Rightarrow\,y-(-5)=4(x-7)$$

$$\Rightarrow\,y+5=4x-28$$

$$\Rightarrow\,4x-y-33=0$$

### Find the equation of tangent line to the graph of  $$y=f(x)$$  at  $$x=7$$  if $$f(7)=-5$$ and $$f'(7)=4$$.

A

$$4x-y-33=0$$

.

B

$$x+2y-1=0$$

C

$$x=7$$

D

$$y=-8$$

Option A is Correct

# Slope of Tangent from the Graph

Consider the graph $$y=f(x)$$ as shown in figure.  Different tangents at points $$P,\,Q,\;R$$ are drawn as shown in figure  We see

At point $$P$$

i.e. at $$x=x_1$$, tangent makes acute angle with $$x-$$ axis which implies

Slope of tangent $$\to\;f'(x_1)>0$$

at $$x=x_1$$

At point $$Q$$

i.e. at $$x=x_2$$, tangent is parallel to $$x-$$ axis which implies

Slope of tangent at $$x=x_2\,\to$$ $$f'(x_2)=0$$

At point $$R$$

i.e. at $$x=x_3$$, tangent makes obtuse angle with $$x-$$ axis which implies

Slope of tangent at $$x=x_3\,\to$$ $$f'(x_3)<0$$

It also concludes

$$f'(x_1)>f'(x_2)>f'(x_3)$$  #### Consider the graph of  $$y=f(x)$$  as shown. Identify the correct statement.

A $$f'(2)<f'(4)<f'(6)$$

B $$f'(2)>f'(4)>f'(6)$$

C $$f'(4)<f'(6)<f'(2)$$

D $$f'(6)<f'(2)<f'(4)$$

×

$$f'(2)=$$ Slope of tangent at $$x=2$$,

$$\Rightarrow\,f'(2)>0$$

(tangent makes an acute angle with $$x$$- axis) $$f'(4)=$$ Slope of tangent at $$x=4$$

$$\Rightarrow\,f'(4)=0$$

(tangent is parallel to $$x$$- axis) $$f'(6)=$$ Slope of tangent at $$x=6$$

$$\Rightarrow\,f'(6)<0$$

(tangent makes an obtuse angle with $$x$$- axis) $$f'(2)>f'(4)>f'(6)$$ ### Consider the graph of  $$y=f(x)$$  as shown. Identify the correct statement. A

$$f'(2)<f'(4)<f'(6)$$

.

B

$$f'(2)>f'(4)>f'(6)$$

C

$$f'(4)<f'(6)<f'(2)$$

D

$$f'(6)<f'(2)<f'(4)$$

Option B is Correct

# Average Rate of Change of a Function

Let $$y=f(x)$$ be a function , if $$x$$ changes from $$x _1$$ to $$x _2$$, the change in $$x$$ (called the increment of $$x$$) is  $$\Delta x=x_2 -x_1$$

The corresponding change in $$y$$ is $$\Delta y = f(x_2) - f(x_1)$$

The difference quotient = $$\dfrac{\Delta y}{\Delta x} = \dfrac{f(x_2)-f(x_1)}{x_2-x_1}$$  is called the average rate of change of $$y$$ with respect to $$x$$ over the interval $$(x_1,x_2)$$

• It can be interpreted as slope of secant line PQ as shown.  #### Let $$f(x) = \dfrac{2x^2-x+1}{3}$$ , find the average rate of change of $$f$$ with respect to $$x$$ over the interval [1,3].

A $$\dfrac{8}{3}$$

B $$\dfrac{7}{3}$$

C $$\dfrac{5}{3}$$

D $$3$$

×

Average rate of change = The difference quotient $$= \dfrac{\Delta y}{\Delta x} = \dfrac{f(x_2)-f(x_1)}{x_2-x_1}$$

In this case , $$x_2 = 3, \,\,x_1= 1$$

$$\Rightarrow f(x_2) = f(3) = \dfrac{2×9-3+1}{3} = \dfrac{16}{3}$$

$$\Rightarrow f(x_1) = f(1) = \dfrac{2×1-1+1}{3} = \dfrac{2}{3}$$

$$\therefore$$ Average rate of change = $$\dfrac{\dfrac{16}{3}-\dfrac{2}{3}}{3-1} = \dfrac{\dfrac{14}{3}}{2}$$

$$= \dfrac{7}{3}$$

### Let $$f(x) = \dfrac{2x^2-x+1}{3}$$ , find the average rate of change of $$f$$ with respect to $$x$$ over the interval [1,3].

A

$$\dfrac{8}{3}$$

.

B

$$\dfrac{7}{3}$$

C

$$\dfrac{5}{3}$$

D

$$3$$

Option B is Correct

# Average Velocity

Average velocity is the average rate of change of position with respect  to time during an interval $$(t_1,t_2)$$ .

Average Velocity = $$\dfrac{\Delta x}{\Delta t} = \dfrac{x(t_2)-x(t_1)}{t_2-t_1}$$

where x=position,t=time  #### Let position of a  particle is given by $$s(t) = 2\,t^2 -t+3,$$ where $$t$$is in seconds and position is in meter . Find the average velocity during the time  interval 1 to 3 second.

A 1 m/s

B 5 m/s

C 8 m/s

D 2 m/s

×

Average Velocity = $$\dfrac{\Delta s}{\Delta t} = \dfrac{s(t_2)-s(t_1)}{t_2-t_1}$$

$$\Rightarrow s(t_2) = s(3) = 2× (3)^2 - 3+3$$

$$=18 \,m$$

$$\Rightarrow s(t_1) = s(1) = 2× 1 - 1+3$$

$$\Rightarrow-1+3=2 \,m$$

$$\therefore$$ Average Velocity  $$= \dfrac{18-2}{3-1}$$

$$= \dfrac{16}{2} = 8 \,m/s$$

### Let position of a  particle is given by $$s(t) = 2\,t^2 -t+3,$$ where $$t$$is in seconds and position is in meter . Find the average velocity during the time  interval 1 to 3 second.

A

1 m/s

.

B

5 m/s

C

8 m/s

D

2 m/s

Option C is Correct