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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