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 \((a,\,f(a))\)

Illustration Questions

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 \(x=2\)

\(=12\)

Illustration Questions

If \(\lim\limits_{h\to 0}\,\left(\dfrac{\sqrt[3]{1+h}-1}{h}\right)= f'(a)\)  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 \(\lim\limits_{h\to 0}\,\left(\dfrac{\sqrt[3]{1+h}-1}{h}\right)= f'(a)\)  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

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 \(f'(a)\).

Illustration Questions

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

\(f'(x_1)>f'(x_2)>f'(x_3)\)

Illustration Questions

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)

image

\(f'(4)=\) Slope of tangent at \(x=4\)

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

 (tangent is parallel to \(x\)- axis)

image

\(f'(6)=\) Slope of tangent at \(x=6\)

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

 (tangent makes an obtuse angle with \(x\)- axis)

image

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

image

Consider the graph of  \(y=f(x)\)  as shown. Identify the correct statement.

image
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

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, \(x=a\).

Illustration Questions

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 \(f'(2)\) \(=10×2^9\)

Illustration Questions

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

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.

Illustration Questions

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

Illustration Questions

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

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