MathCity.org Exercise 2.6 (Solutions)



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Exercise 2.6 (Solutions)

Calculus and Analytic Geometry, MATHEMATICS 12 Available online @ , Version: 3.0

2.10 Derivative of General Exponential Function (Page 80)

A function define by f (x) = ax where a > 0, a 1

is called general exponential function.

Suppose

y = ax

y + y = ax+ x y = ax+ x - y

y = ax+ x - ax

Since y = ax

y = ax (a x -1)

Dividing by x

y x

=

ax (a x -1) x

Taking limit as x 0

lim

x0

y x

=

lim

x0

a

x

(a x x

-1)

dy dx

=

lim

x0

a

x

a x

-1 x

dy dx

=

ax

lim

x0

a x

-1 x

d (ax) = ax.ln a

dx

Since lim ax -1=ln a x0 x

Derivative of Natural Exponential Function The exponential function f (x) = ex , where e = 2.71828..., is called Natural

Exponential Function.

Suppose

y = ex

Do yourself ... Just Change a by e in above article. You'll get

d ex = ex dx

2.11 Derivative of General Logarithmic Function (page 81)

If a > 0, a 1 and x = a y , then the function defined by y = loga x (x > 0) is

called General Logarithmic Function.

Suppose

y = loga x y + y = loga (x + x) y = loga (x + x) - y

y = loga (x + x) -loga x

=

loga

x + x x

Since

loga

m - loga

n = loga

m n

Dividing both sides by x

y x

=

1 x

loga

x

+ x

x

FSc-II / Ex- 2.6 - 2



Taking limit as x 0

lim

x0

y x

=

lim

x0

1 x

loga

x+ x

x

dy dx

=

lim

x0

1 x

loga

1+

x x

=

lim

x0

x x

1 x

loga

1+

x x

dy dx

=

1 x

lim

x0

x x

loga

1+

x x

x

dy dx

=

1 x

lim

x0

log

a

1+

x x

x

dy dx

=

1 x

loga

lim

x0

1

+

x x

x

x

dy dx

=

1 x

loga

e

d dx

(

log

a

x)

=

1 x

1 loge a

d dx

(

log

a

x)

=

x

1 ln a

?ing and ?ing by x

Since mloga x =loga xm

1

Since lim(1+ x)x =e x0

Since

loga

b= 1 logb

a

Since loge a =ln a

Derivative of Natural Logarithmic Function

The logarithmic function f (x) = loge x where e = 2.71828...is called Natural

Logarithmic Function. And we write ln x instead of loge x for our ease.

Suppose

y = ln x

y + y = ln (x + x) y = ln (x + x) - y

y = ln (x + x) -ln x

y

=

ln

x

+ x

x

Since

ln

m

-

ln

n

=

ln

m n

=

ln 1+

x x

Dividing both sides by x

y x

=

1 x

=

ln 1+

x x

Taking limit as x 0

lim

x0

y x

=

lim

x0

1 x

ln

1+

x x





dy dx

=

lim

x0

x x

1 x

ln

1+

x x

dy dx

=

1 x

lim

x0

x x

ln

1

+

x x

x

dy dx

=

1 x

lim

x0

ln

1+

x x

x

dy dx

=

1 x

ln

lxim0

1

+

x x

x

x

dy dx

=

1 x

ln

e

d dx

(ln

x)

=

1 x

1

d (ln x) = 1

dx

x

FSc-II / Ex- 2.6 - 3 ?ing and ?ing by x

Since mln x = ln xm

1

Since lim(1+ x)x =e x0

Since ln e = loge e =1

Question # 1 Find f (x) if

Exercise 2.6 (Questions)

(i) f ( x) = e x-1

1

(ii) f ( x) = x3e x ,( x 0)

(iii) f (x) = ex (1 + ln x)

(iv)

f (x)

=

ex e-x +1

( ) (v) f (x) = ln ex + e-x

(vi)

f (x)

=

eax - e-ax eax + e-ax

( ) ( ) (vii) f (x) = ln e2x + e-2x (viii) f (x) = ln e2x + e-2x

Solution

(i) f (x) = e x-1

Diff. w.r.t x

d f (x) = d e x-1

dx

dx

( ) f (x) = e x-1 d x -1 dx

=

e

x -1

1

-1

x2

- 0

= e x -1

Ans.

2

2x

1

(ii) f (x) = x3e x

Diff. w.r.t x



FSc-II / Ex- 2.6 - 4



d

f (x) =

d

1

x3e x

dx

dx

f (x) =

x3

d

11

ex +ex

d

x3

dx

dx

( ) =

1

x3e x

d dx

1 x

1

+ex

3x2

( ) =

1

x3e x

-

1 x2

+

1

ex

3x2

1

1

1

= - xe x + 3x2 e x = xe x (3x -1)

d dx

1 x

=

d dx

x-1

=

- x -2

=

-

1 x2

Ans.

(iii) f (x) = ex (1+ ln x)

Diff. w.r.t x

d f (x) = d ex (1+ ln x)

dx

dx

f (x) = ex d (1+ ln x) +(1+ ln x) d ex

dx

dx

=

ex

0

+

1 x

+

(1

+

ln

x)

ex

f (x)

=

ex

1 x

+1+

ln

x

or

f (x)

=

ex

1

+

x

(1

+

ln

x)

x

(iv)

f (x)

=

ex e-x +1

Diff. w.r.t x

d f (x) dx

=

d ex

dx

e-x

+

1

( ) ( ) e-x +1 d ex - ex d e-x +1

f (x) =

dx

dx

( ) e-x +1 2

( ) ( ) e-x + 1 ex - ex e-x (-1) + 0

( ) ex e-x + 1 + e-x

=

=

( ) e-x + 1 2

( ) e-x + 1 2

(( ) )

f (x) =

ex 2e-x + 1 e-x +1 2

Ans.





FSc-II / Ex- 2.6 - 5

( ) (v) f (x) = ln ex + e-x

Diff. w.r.t x

( ) d f (x) = d ln ex + e-x

dx

dx

( ) f (x) =

1

d ex + e-x

( ) ex + e-x dx

( ) =

1

ex + e-x (-1)

( ) ex + e-x

f (x) =

ex - e-x ex + e-x

or

f (x) = tanh x

(vi)

f (x)

=

eax - e-ax eax + e-ax

Diff. w.r.t x

d f (x) dx

=

d dx

eax eax

- +

e-ax e - ax

tanh x =

ex - e-x ex + e-x

( ) ( ) ( ) ( ) eax + e-ax d eax - e-ax - eax - e-ax d eax + e-ax

=

dx

dx

( ) eax + e-ax 2

( )( ) ( )( ) eax + e-ax eax (a) - e-ax (-a) - eax - e-ax eax (a) + e-ax (-a)

=

( ) eax + e-ax 2

( )( ) ( )( ) a eax + e-ax eax + e-ax - a eax - e-ax eax - e-ax

=

( ) eax + e-ax 2

( ) ( ) =

a

eax

+ e-ax

2

-

eax - e-ax

2

( ) eax + e-ax 2

= a (e2ax + e-2ax + 2eaxe-ax ) - (e2ax + e-2ax - 2eaxe-ax )

( ) eax + e-ax 2

= a e2ax + e-2ax + 2 - e2ax - e-2ax + 2

( ) eax + e-ax 2

eaxe-ax = e0 = 1

( ) f (x) =

4a eax + e-ax 2

Ans.



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