Download Derivative of General Exponential and Logarithmic

derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Derivatives Involving ln x
MCV4U: Calculus & Vectors
Recap
Determine any value(s) of x where the tangent to
f (x) = 2x · ln3 x is horizontal.
Derivatives of General Exponential and
Logarithmic Functions
Use the product and chain rules to find the derivative.
f 0 (x) = 2 ln3 x + 2x · 3 ln2 x ·
1
x
= 2 ln3 x + 6 ln2 x
J. Garvin
= 2 ln2 x (ln x + 3)
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 2/11
Slide 1/11
derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Derivatives Involving ln x
Derivatives of Exponential Functions
Since the tangent is horizontal, the slope is zero. Solve each
factor.
To find the derivative of a general exponential function,
f (x) = b x where b 6= e, we can a technique similar to that
used to find the derivative of f (x) = ln x.
2 ln2 x = 0
ln x + 3 = 0
2
y = bx
ln x = −3
ln x = 0
ln y = ln b x
x = e −3
ln x = 0
x =1
x=
1
e3
1
y
Therefore, the tangent is horizontal when x = 1 or when
x = e13 .
·
dy
dx
dy
dx
= ln b · x
= ln b
= y ln b
= b x ln b
Derivative of y = b x
If f (x) = b x , then f 0 (x) = b x ln b. If y = b x ,
dy
dx
derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Derivatives of Exponential Functions
Derivatives of Exponential Functions
Example
Example
Determine the derivative of f (x) = 5x .
Determine the slope of the tangent to y =
This is pretty straightforward.
dy
dx
Example
Determine the derivative of f (x) = 3x 2 · 2x .
Use the product rule.
x
2
x
f (x) = 6x · 2 + 3x · 2 ln 2
= 3x · 2x (2 + x ln 2)
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 5/11
2x
when x = 4.
x
Use the quotient rule.
f 0 (x) = 5x ln 5
0
= b x ln b.
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 4/11
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 3/11
Thus,
dy dx x=4
=
2x ln 2 · x − 2x
x2
2x (x ln 2 − 1)
=
x2
=
24 (4 ln 2 − 1)
= 4 ln 2 − 1.
42
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 6/11
derivatives of trigonometric, exponential & logarithmic functions
derivatives of trigonometric, exponential & logarithmic functions
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Using implicit differentiation, we can also determine the
derivative of a general logarithmic function.
Example
Determine the derivative of f (x) = 6 log2 x + 5x.
Recall that if y = logb x, then b y = x.
Use the constant multiple and sum rules for derivatives.
6
f 0 (x) =
+5
x ln 2
ln b y = ln x
y ln b = ln x
ln b ·
dy
dx
dy
dx
=
1
x
=
1
x ln b
Example
Determine the slope of the tangent to y = log(3x + 2) when
x = 1.
Derivative of y = logb x
If f (x) = logb x, then
f 0 (x)
=
1
x ln b .
If y = logb x,
dy
dx
=
1
x ln b .
x=1
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 8/11
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 7/11
derivatives of trigonometric, exponential & logarithmic functions
Derivatives of Logarithmic Functions
derivatives of trigonometric, exponential & logarithmic functions
Derivatives of Logarithmic Functions
Example
Example
At what point on the function f (x) = log5 x is the slope
The derivative is f 0 (x) =
1
x ln 5 .
Set this equal to
1
25
1
x ln 5
25
ln 5
=
x=
f (x) = log5
Assuming log x is the common logarithm (base 10) of x, the
3
derivative is dy
dx = (3x+2) ln 10 . Note the chain rule here.
Therefore, dy
= 5 ln3 10 .
dx 25
ln 5
= 2 − log5 (ln 5).
1
25 ?
1
25 .
1
25
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 9/11
derivatives of trigonometric, exponential & logarithmic functions
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 11/11
Using the chain rule, dy
dx = 2 log3 x ·
dy Thus, dx = 2 log3 27 · 27 1ln 3 =
x=27
Therefore, the function
has a slope of
at the point
25
ln 5 , 2 − log5 (ln 5) , or approximately (15.5, 1.7).
Questions?
2
Determine the equation of the line parallel to y = (log
3 x)
when x = 27, if it passes through the point 6, ln53 .
5
ln 3
5
ln 3
=
=
b=
1
x ln 3 .
2
9 ln 3 .
2
9 ln 3 (6) +
4
3 ln 3 + b
11
3 ln 3
The equation of the line is y =
2
9 ln 3 x
J. Garvin — Derivatives of General Exponential andLogarithmic Functions
Slide 10/11
b
+
11
3 ln 3 .