MATHEMATICS 0110A
DERIVATIVES OF GENERAL EXPONENTIAL AND LOGARITHMIC
FUNCTIONS
d
d
[ex ] = ex and dx
[ln x] = x1 . But what about exponential and logaWe have seen that dx
rithmic functions with other bases? How can we find the derivatives of those? Using what
we have learned so far there are, in each case, 2 different approaches we can use to find a
formula for the derivative of a general exponential or a general logarithmic function. We
can either change the base to e, or else use logarithmic differentiation (for exponential functions) or implicit differentiation (for logarithmic functions). Both approaches are shown here.
The derivative of f(x) = bx
Approach 1: Changing the base
Suppose that y = bx . Then
x
y = eln b
so
y = ex ln b
Therefore
d
d x ln b dy
x ln b
e
=e
=
(x ln b) = ex ln b ln b
dx
dx
dx
(because of course ln b is just a constant). But ex ln b = bx so we get
dy
= bx ln b
dx
Approach 2: Using Logarithmic Differentiation
Let y = bx and take natural logarithms of both sides of this equation:
ln y = ln bx
so
ln y = x ln b
and now we differentiate:
y′
d
=
(x ln b) = ln b
y
dx
(since ln b is just a constant). We multiply through by y to get y ′ = y ln b and use the fact
that y = bx , to get y ′ = bx ln b.
Either way, we see that:
For f (x) = bx, the derivative is f ′(x) = bx ln b
d
d
1 x
1 x
d
For instance, dx
[2x ] = 2x ln 2, dx
=
ln 13 and of course dx
(ex ) = ex ln e = ex .
3
3
As always, we use this new rule in conjunction with all the other differentiation rules.
2
Mathematics 0110A
Derivatives General Exp. and Log. Functions
Example 1
Find the derivative of f (x) = 3x (2x − x3 )
We use the product rule in the usual way:
f ′ (x) = (3x ln 3)(2x − x3 ) + 3x (2 − 3x2 ) = 3x [2 − 3x2 + (ln 3)(2x − x3 )]
Example 2
Differentiate y =
x2 + x3
2(5x ) − 1
Of course we use the quotient rule for this:
(2x + 3x2 )[2(5x ) − 1] − (x2 + x3 )[2(5x ln 5)]
dy
=
dx
(2(5x ) − 1)2
Example 3
4x
at the point on the curve with x = 0.
e2x − x
(4x ln 4)(e2x − x) − 4x (e2x (2) − 1)
We get
f ′ (x) =
(e2x − x)2
so the slope of the tangent line is
Find the slope of the tangent line to f (x) =
f ′ (0) =
(ln 4)(1) − (2 − 1)
(40 ln 4)(e2(0) − 0) − 40 (e2(0) (2) − 1)
=
= (ln 4) − 1
2(0)
2
(e
− 0)
(1)2
Of course, if we have an exponent that is not just x, but is some function of x, i.e., if we
have f (x) = bg(x) then we use the Chain Rule:
For f (x) = bg(x) , the derivative is f ′(x) = (bg(x) ln b)g ′(x)
Example 4
If f (x) = 52x + 25x , find f ′ (0) and f ′ (−1).
f (x) = 52x + 25x
d
d
′
2x
5x
so f (x) = 5 (ln 5)
(2x) + 2 (ln 2)
(5x) = 52x (2 ln 5) + 25x (5 ln 2)
dx
dx
We get f ′ (0) = (50 )(2 ln 5) + (20 )(5 ln 2) = (2 ln 5) + (5 ln 2)
so f ′ (0) = (ln 52 ) + (ln 25 ) = ln(52 × 25 ) = ln(25 × 32) = ln(25 × 4 × 8) = ln 800
2 ln 5 5 ln 2
2 ln 5 5 ln 2
+
and f ′ (−1) = 5−2 (2 ln 5) + (2−5 (5 ln 2) = 2 + 5 =
5
2
25
32
Mathematics 0110A
Derivatives General Exp. and Log. Functions
3
Example 5
dy
if xexy = y2x+y .
Find dx
We don’t have the form y = f (x), so we need to use implicit differentiation:
xexy = y2x+y →
dy
exy + xexy 1y + x dx
→
exy + xyexy + x2 exy
→
→
dy
dx
d
1exy + x exy dx
(xy)
=
→
→ x2 exy
d
[xexy ]
dx
dy
dx
− 2x+y
dy
dx
dy
dx
dy
dx
dy
− y2x+y (ln 2) dx
=
d
[y2x+y ]
dx
(product rule on both sides)
d
2x+y + y (2x+y ln 2) dx
(x + y)
dy
dx
= 2x+y
dy
dx
+ y2x+y (ln 2) 1 +
= 2x+y
dy
dx
dy
+ y2x+y (ln 2) + y2x+y (ln 2) dx
(chain rule)
= y2x+y (ln 2) − exy − xyexy
[x2 exy − 2x+y (1 + y ln 2)] = y2x+y (ln 2) − exy (1 + xy)
dy
dx
→
=
y2x+y (ln 2) − exy (1 + xy)
x2 exy − 2x+y (1 + y ln 2)
The derivative of f(x) = logb x
Approach 1: Using the Change of Base formula
ln x
We know that logb x =
, so we get
ln b
d
1
1
d
1
1
1
d ln x
′
=
ln x =
(ln x) =
× =
f (x) =
dx ln b
dx
ln b
ln b
dx
ln b x
x ln b
Approach 2: Using Implicit Differentiation
Let y = logb x, so that x = by . We use implicit differentiation:
d
d y
dy
(x) =
(b )
→
1 = (by ln b)
→
dx
dx
dx
Either way, we see that:
dy
1
1
= y
=
dx
b ln b
x ln b
For f (x) = logb x, the derivative is f ′(x) =
So for instance,
the usual rules.
d
(log2
dx
x) =
1
x ln 2
and
d
(log27
dx
x) =
1
.
x ln 27
1
x ln b
Again, we use this along with all
Example 6
Differentiate y = x2 (x + 2 log5 x).
2x2
1
dy
2x
2
2
2
= 2x +4x log5 x+x +
= 2x(x+2 log5 x)+x 1 + 2
= 3x2 +4x log5 x+
dx
x ln 5
x ln 5
ln 5
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Mathematics 0110A
Derivatives General Exp. and Log. Functions
Example 7
Let f (x) =
x log2 x
′ 1
.
Find
f
.
2
4 − 4x
1
(4 − 4x ) − (x log2 x)(−4x ln 4)
1(log
x)
+
x
2
x ln 2
f ′ (x) =
(4 − 4x )2
log2 x + ln12 (4 − 4x ) + x4x (ln 4) log2 x
=
(4 − 4x )2
log2 21 + ln12 (4 − 41/2 ) + 21 41/2 (ln 4) log2 12
1
′
=
so f
2
2
(4 − 41/2 )
√
√
log2 2−1 + ln12 (4 − 4) + 21
4(ln 22 )(log2 2−1 )
√
=
(4 − 4)2
1
− 1 (2) + (2 ln 2)(−1)
ln 2
=
(2)2
1
1 ln 2
=
− −
2 ln 2 2
2
And of course, we could have the logarithm of a function of x, rather that just x, in which
case we again use the Chain Rule:
g ′(x)
For f (x) = logb g(x), the derivative is f ′(x) =
g(x)(ln b)
Example 8
Find the derivative of f (x) = log2 x3 + log5 3x .
Approach 1:
Both terms involve the logarithm of a function of x, so we need the Chain rule.
d 3
d x
1
1
′
(x ) +
(3 )
f (x) =
x3 ln 2
dx
3x ln 5
dx
1
3
1
ln 3
3
2
x
=
(3x
)
+
(3
ln
3)
=
+
=
+ log5 3
x3 ln 2
3x ln 5
x ln 2 ln 5
x ln 2
Approach 2:
We can use the properties of logarithms to simplify before differentiating. We have:
f (x) = log2 x3 + log5 3x = 3 log2 x + x log5 3
Now we can differentiate without the chain rule:
1
3
′
f (x) = 3
+ 1(log5 3) =
+ log5 3
x ln 2
x ln 2
(Remember, log5 3 is just a constant.)
Mathematics 0110A
Derivatives General Exp. and Log. Functions
5
Example 9
If y = log2 (log3 (5x)), find y ′ .
This time, we certainly need the chain rule, and more than once. We get:
d
y′ =
[log2 (log3 (5x))]
dx
1
d
=
(log3 (5x))
[log3 (5x)] ln 2
dx
1
1
d
=
(5x)
[log3 (5x)] ln 2
5x ln 3 dx
1
1
(5)
=
[log3 (5x)] ln 2
5x ln 3
5
=
5x(ln 2)(ln 3)(log3 (5x))
1
=
x(ln 2)(ln 3)(log3 (5x))
Notice: We can also express what we did here by introducing new variables. We start from
the inside and work out. We have y = log2 (log3 (5x)):
du
Let u = 5x so that
= 5.This gives y = log2 (log3 u).
dx
dv
1
dv
dv du
Now let v = log3 u so that
=
. Of course,
=
×
du
u ln 3
dx
du dx
1
dy
=
and so
Then y = log2 v with
dv
v ln 2
dy dv
dy dv du
dy
=
×
=
×
×
dx
dv dx
dv du dx
We get
5
1
5
1
dy
(5) =
=
=
dx
v ln 2
u ln 3
[(log3 u)(ln 2)](u ln 3)
(log3 (5x))(ln 2)(5x ln 3)
which we can simplify and rearrange as before.
6
Mathematics 0110A
Derivatives General Exp. and Log. Functions
Homework Exercises
(1) Differentiate each of the following:
(a) f (x) = 5x
(e) f (x) =
(b) f (x) = 2(3x )
6x +x
2x−4
(f ) h(t) =
t(2t −2t)
1+t2
(c) g(x) = 2x + 3x + ex (d) y = x2 4x
(g) f (x) =
√
(h) y = ln(x + 4x )
4x
(2) Find f ′ (x) in each of the following.
(a) f (x) = 32x
(e) f (x) = x2 2(x
(b) f (x) = 23x
2)
(f ) f (x) =
(c) f (x) = 4x/2
4x2 35x−1
2x −x
(d) f (x) = 5x
2 +3x+2
2
(g) f (x) = ln(x + 3(x ) )
(3) Find the derivative of each of the following functions.
(a) f (x) = log2 x
(b) g(x) = x + 2 log3 x
(d) f (x) = x log x
(e) y =
(4) Find
dy
dx
(c) h(x) =
ex −x2
log3 x
(f ) s =
p
√
x + log5 x
t2 + 4 log3 t
in each of the following:
(a) y = log5 (2x + ex )
(b) y = ex + log2 (5x)
(c) y = 2log x
(d) y = x2 + x2 log x2
(e) y = x log(ln x)
(f ) y = log2 (log5 (log x))
(g) xby = x2 − y 2
(h) log(x2 y + xy 2 ) = x log e
(5) Find the slope of the tangent line to y = f (x) at x = a for each of the following.
(a) f (x) = 3x ; a = log3 2
(b) f (x) = 2x
2 +e2
; a=e
(c) f (x) = 2 1+
(6) If y = xlog2 x , find y ′.
(7) Evaluate the following limits:
log 15 + h − log 51
23+h − 23
(a) lim
(b) lim
h→0
h→0
h
h
Hint: Recognize the form of the limit. What does it represent?
√
x
; a=4