Derivatives of Exponential and Log Functions

These 2 are the easiest to differentiate and yet so many students find them confusing.

In all these expressions, u is some f (x), so du is the derivative of f (x).

Reminders:

The base of an exponential function (e ^x or a ^x) is POSITIVE!

the base of the natural exponential function is e = 2.718. . .

Negative exponents create fractions, not negative values! so

Exponential Functions:

Note 1: exponential function: y ' = (derivative of exponent) (original function unchanged).
that's du e ^u -- we take derivative of power function (up top) and plunk it down front.

So if y = e ^{5 x – 3} then y ' = 5 e ^{5 x – 3} .

See it? The derivative of (5x – 3 -- the exponent) multiplying the original function.

Note 2: Notice the difference between the derivatives of y = e ^u and y = a ^u. There's no ln a in the derivative of e ^u. Well, actually, there is. However, since a = e, ln a = ln e which = 1, so we don't bother to write it.

Examples:

Example 1:

, so ,

Example 2:

y = ( 3 ) ^{2 x – 5}, so y ' = 2(3) ^{2 x – 5} ln 3

Example 3: a product

f (x) = x³ ( e^{4x³ – 6x} ) -- f '(x) = 3x²( e^{4x³ – 6x} ) + x³(12x² – 6)( e^{4x³ – 6x} ).

Example 4: a quotient

Log Functions:

Reminders:

ln x means log _e x -- it's pronounced lawn, the base is e = 2.718. . .
and in log _a x, "a" is a base other than e. Also, log x means log base 10 of x, so a = 10.

In all these expressions, u is some f (x), so du is the derivative of f (x).

function	derivative	function	derivative
(ln u)		(log a u)

Note 3: logarithmic function: if y = ln (thing) then

So if y = ln (5x ³ – 4x ² + 3x) then

Note 4: Notice the difference between the derivatives of y = ln u and y = log _a u. Same as in Note2 -- the ln a is missing in the denominator when the base = e because ln e = 1.

Examples:

Example 1:

y = – 3 ln (x ⁶ – 2x ² + 9), so

Example 2:

y = log ₃ (4x ³ – 6x + 1), so

Example 3: a product

y = (7x ⁴ – 5x) ln (4x ³ – x ²), so

Reminders:

Logs are exponents so we can apply the rules of exponents to log expressions:

1) log c (MN) = log c M + log c N. so log c 2y = log c 2 + log c y	log of a product = sum of the logs. note: sum of logs not log of the sum!
2) so	log of a quotient = difference of the logs.
3) log c (M) n = n log c M so log c x 23 = 23 log c x	log of M to a power n = power n times log M
4) log c1 = 0	the zero power of any base = 1

Reminders:

the exponential form of log _c x = a is c ^a = x .

the log form of c ^a = x is log _c x = a .

e ^{ln x} = x and ln e ^x = x

These are useful when solving exponential and log equations.

Example 4: applying the rules of logs to a messy function before differentiating.

Example 5:

y = ln [ ln (4x² + 3) ] so the derivative is

In this case u is ln (4x² + 3) and we're taking ln of it, so du/u is as shown.

Example 6: finding equation of a tangent

Find the equation of the tangent to y = (x – 1)e ^x + 3 ln x + 2 at the point P(1, 2).

We know the point so we need to find the slope of the tangent at P. Tangent slope = y ' .
The first term in the function is a product, so we use the product rule.

so the slope of this tangent = e + 3

The equation of the tangent therefore is or y = (e + 3) x – (e + 1).

Practice

Differentiate and Simplify

1) y = 2e 7x + 1

2) y = – 6 e – x ³ + 4

3) y = – (5) 7 – 3 x

4) y = 4 (½) x ² + 3x + 6

5) y = 7 ln (10x3 – x + 9)

6) y = ln (sin 2 5x)

7) y = – 3 log 5 (10x3 – x + 9)

8) y = log 7 (cos 5 3x)

9) Find an equation for the tangent to y = x – e ^x that is parallel to 6x – 2y – 7.

10) Find the critical points, increasing and decreasing intervals for y = (1 – ln x )².

Solutions

1) y = 2e 7x + 1 y / = 2(7)(e 7 x + 1)

2) y = – 6 e – x ³ + 4 y / = 18 x ² ( e – x ³ + 4 )

3) y = – (5) 7 – 3 x y ' = 3 (5) 7 – 3 x ln 5

4) y = 4 (½) x ² + 3x + 6 y / = 4(2x + 3) (½) x ² + 3x ln(½)

5) y = 7 ln (10x3 – x + 9)

6) y = ln (sin 2 5x)

7) y = – 3 log 5 (10x3 – x + 9)

8) y = log 7 (cos 5 3x)

9) Find an equation for the tangent to y = x – e ^{– x} that is parallel to the line 6x – 2y – 7 = 0.

Here, we need to find the point on y that makes y' = slope of the line.

Since the slope of Ax + By + C = 0 is – A/B, we need y' = 3.

y' = 1 + e ^{– x} so we must solve 1 + e ^{– x} = 3

this becomes e ^{– x} = 2. Now we apply ln to both sides to get – x = ln 2.

So, x = – ln 2 = ln 2 ^{– 1} = ln (½)

This makes y = ln (½) – e ^{ln 2} = ln (½) – 2.

The equation of the tangent is

this becomes y = 3x – 2 ( ln (½) + 1 )

10) Find the critical points, increasing and decreasing intervals for y = (1 – ln x )².

Note: the domain of this function is x > 0 -- since ln x apllies only to positive Reals.

We find critical points by setting y ' = 0
so we must solve y ' = 2 (1 – ln x )( –1/x ) = 0.

We rewrite the fraction as and now we see that
(1 – ln x ) must = 0 to make the numerator = 0.
From this we get ln x = 1, so x = e. Since y = (1 – ln x )², y = (1 – 1)² = 0.

The critical point is at ( e, 0 ).

Decreasing intervals correspond to y ' < 0 which means 0 < x < e

Increasing intervals correspond to y ' > 0 which means x > e,

Since the curve decreases first, then increases, the critical point at ( e, 0 ) is a minimum.

( Calculus I MathRoom Index )

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