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6.3: Differentiation and Integration of Logarithmic and Exponential Functions

Created by: CK-12

Learning Objectives

A student will be able to:

  • Understand and use the rules of differentiation of logarithmic and exponential functions.
  • Understand and use the rules of integration of logarithmic and exponential functions.

In this section we will explore the derivatives of logarithmic and exponential functions. We will also see how the derivative of a one-to-one function is related to its inverse.

The Derivative of a Logarithmic Function

Our goal at this point to find an expression for the derivative of the logarithmic function y = \log_b x. Recall that the exponential number e is defined as

 e = \lim_{a \to \ 0 }(1 + a)^{1/a}

(where we have substituted a for x for convenience). From the definition of the derivative of f(x) that you already studied in Chapter 2,

 f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)} {h} = \lim_{w \to \ x } \frac {{f(w)}-{f(x)}}{w-x}.

We want to apply this definition to get the derivative to our logarithmic function y= \log_b x. Using the definition of the derivative and the rules of logarithms from the Lesson on Exponential and Logarithmic Functions,

\frac {d}{dx} \left [ {\log_b{x}} \right ] & = \lim _{w \to \ x } \frac {{\log_b{w}}-{\log_b{x}}}{w-x}\\& = \lim _{w \to \ x } \frac {\log_b(w/x)}{w-x}\\& = \lim _{w \to \ x } \left [ \frac{1}{w-x} \log_b \left (\frac{w}{x}\right ) \right ]\\& = \lim _{w \to \ x } \left [ \frac{1}{w-x} \log_b \left ( \frac{x+(w-x)}{x} \right ) \right ]\\& = \lim _{w \to \ x } \left [ \frac{1}{w-x} \log_b \left  (1+ \frac{w-x}{x} \right ) \right ]\\& = \lim _{w \to \ x } \left [ \frac{1} {x(w-x)} \log_b \left  (1+ \frac{w-x}{x} \right ) \right ]\\& = \lim _{w \to \ x } \left [ \frac{x} {x(w-x)} \log_b \left  (1+ \frac{w-x}{x} \right ) \right ].

At this stage, let a =(w-x)/(x), the limit of w\rightarrow x then becomes a\rightarrow 0. Substituting, we get

& = \lim _{a \to \ 0 } \left [ \frac{1}{x} \frac {1}{a} \log_b (1+a)\right ]\\& =\frac {1}{x} \lim _{a \to \ 0 } \left [ \frac{1}{a} \log_b (1+a)\right ]\\& =\frac {1}{x} \lim _{a \to \ 0 } \left [\log_b (1+a)^{1/a} \right ].

Inserting the limit,

 =\frac {1}{x} \log_b \left [\lim _{a \to \ 0 } (1+a)^{1/a} \right ].

But by the definition e = \lim_{a \to \ 0 }(1+a)^{1/a},

 \frac {d}{dx} \left [ \log_b {x}  \right ]= \frac {1}{x} {\log_b{e}}.

From the box above, we can express \log_b{e} in terms of natural logarithm by the using the formula \log_b{w} = \ln{w}/\ln{b}. Then

\log_b{e} = \frac {\ln e}{\ln b} = \frac {1}{\ln b}.

Thus we conclude

\frac{d}{dx}\left [ \log_b {x} \right ]= \frac {1}{x \ln b} > 0,

and in the special case where b = e,

\frac{d}{dx}\left [ { \ln}{x} \right ]= \frac {1}{x} > 0.

To generalize, if u is a differentiable function of x and if u(x) > 0, then the above two equations, after the Chain Rule is applied, will produce the generalized derivative rule for logarithmic functions.

Derivatives of Logarithmic Functions

 \frac {d}{dx} \left [ {\log_b {u}} \right ] & = \frac {1}{u \ln b} \frac {du}{dx}\\\frac {d}{dx} \left [ {\ln u } \right ] & = \frac {1}{u} \frac {du}{dx} = \frac {u'}{u}

Remark: Students often wonder why the constant e is defined the way it is. The answer is in the derivative of f(x) = \ln x. With any other base the derivative of f(x) = \log_b x would be equal  f'(x) = \frac {1}{x \ln b}, a more complicated expression than 1/x. Thinking back to another unexpected unit, radians, the derivative of f(x) = \sin(x) is the simple expression f'(x) = \cos(x) only if x is in radians. In degrees, f'(x) = \frac {\pi}{180} \cos(x), which is more cumbersome and harder to remember.

Example 1:

Find the derivative of y = \ln(2x^2 - 4x + 3).

Solution:

Since  \frac {d}{dx} \left [ {{ \ln}u} \right ] = \frac {1}{u} \frac {du}{dx} , for u = 2x^2 - 4x + 3,

\frac {dy}{dx} & = \frac {1}{2x^2-4x+3} \frac {d}{dx} \left [ {2x^2-4x+3} \right ]\\& = \frac {1}{2x^2-4x+3} (4x-4)\\& = \frac {4(x-1)}{2x^2-4x+3}.

Example 2:

Find  \frac {d}{dx} \left [ {\ln(\sin x)} \right ].

Solution:

\frac {d}{dx} \left [ {\ln(\sin x)} \right ] & = \frac {1}{\sin x} \cdot \left [ {\cos x} \right ]\\& = \frac {\cos x}{\sin x}\\& = \cot x.

Example 3:

Find  \frac {d}{dx} \left [ {\ln(\cos 5x)^3} \right ].

Solution:

Here we use the Chain Rule:

\frac {d}{dx} \left [ {\ln(\cos 5x)^3} \right ] & = \frac {1}{\cos^3 {5x}} \cdot \left [ 3(\cos 5x)^2 \cdot (-\sin 5x) \cdot (5) \right ]\\& = \frac {1}{\cos^3 {5x}} \cdot \left [ {-15 \cos^2 {5x}} \cdot {\sin {5x}} \right ]\\& = \frac {-15 \sin {5x}} {\cos {5x}}\\& = {-15 \tan {5x}}.

Example 4:

Find the derivative of y = x^3 \log_5 2x.

Solution:

Here we use the Product Rule along with  \frac{d} {dx} [\log_b u] = \frac{1} {u \ln b} \frac{du} {dx}:

\frac{d} {dx} [x^3 \log_5 2x] & = x^3 \cdot \frac{d} {dx} [\log_5 2x] + \frac{d} {dx} [x^3] \cdot \log_5 2x\\& = x^3 \cdot \frac{1} {x \ln 5} + 3x^2 \cdot \log_5 2x\\& = \frac{x^2} {\ln 5} + 3x^2 \log_5 2x.

Example 5:

Find the derivative of  y = \ln \frac{x} {x + 1} \cdot

Solution:

We use the Quotient Rule and the natural logarithm rule:

y' & =  \frac{1} {\frac{x} {x+1}} \cdot \frac{(x + 1)(1) - (1)(x)} {(x + 1)^2}\\& = \frac{x + 1} {x} \cdot \frac{1} {(x + 1)^2}\\& = \frac{1} {x(x + 1)}.

Integrals Involving Natural Logarithmic Function

In the last section, we have learned that the derivative of y = \ln u(x) is  y' = \frac{1} {u(x)} . u'(x). The antiderivative is

 \int \frac{u'(x)} {u(x)} dx = \ln |u(x)| + C.

If the argument of the natural logarithm is x, then  \frac{d} {dx} [\ln x] = 1/x, thus

 \int \frac{1} {x} dx = \ln |x| + C.

Example 6:

Evaluate  \int \frac{1} {x + 1} dx.

Solution:

In general, whenever you encounter an integral with an integrand as a rational function, it might be possible that it can be integrated with the rule of natural logarithm. To do so, determine the derivative of the denominator. If it is the numerator itself, then the integration is simply the \ln of the absolute value of the denominator. Let’s test this technique.

 \int \frac{1} {x + 1} dx.

Notice that the derivative of the denominator is 1, which is equal to the numerator. Thus the solution is simply the natural logarithm of the absolute value of the denominator:

 \int \frac{1} {x + 1} dx = \ln |x + 1| + C.

The formal way of solving such integrals is to use u-substitution by letting u equal the denominator. Here, let u = x + 1, and du = dx. Substituting,

\int \frac{1} {x + 1} dx &= \int \frac{1} {u} du\\&= \ln |u| + C\\&= \ln |x + 1| + C.

Remark: The integral must use the absolute value symbol because although x may have negative values, the domain of \ln(x) is restricted to x \ge 0.

Example 7:

Evaluate  \int \frac{4x + 1} {4x^2 + 2x + 1} dx.

Solution:

As you can see here, the derivative of the denominator is 8x + 2. Our numerator is 4x + 1. However, when we multiply the numerator by 2, we get the derivative of the denominator. Hence

\int \frac{4x + 1} {4x^2 + 2x + 1} dx &= \frac{1} {2} \int \frac{2(4x + 1)} {4x^2 + 2x + 1} dx\\&= \frac{1} {2} \int \frac{8x + 2} {4x^2 + 2x + 1} dx\\&= \frac{1} {2} \ln|4x^2 + 2x + 1| + C.

Again, we could have used u-substitution.

Example 8:

Evaluate  \int \tan x dx.

Solution:

To solve, we rewrite the integrand as

 \int \tan x dx = \int \frac{\sin x} {\cos x} dx.

Looking at the denominator, its derivative is  -\sin x. So we need to insert a minus sign in the numerator:

& = - \int \frac{-\sin x} {\cos x} dx\\& = -\ln |\cos x| + C.

Derivatives of Exponential Functions

We have discussed above that the exponential function is simply the inverse function of the logarithmic function. To obtain a derivative formula for the exponential function with base b, we rewrite y = b^x as

x = \log_b  y.

Differentiating implicitly,

1 = \frac{1} {y \ln b} \cdot \frac{dy} {dx}.

Solving for  \frac{dy} {dx} and replacing y with b^x,

\frac{dy} {dx} = y \ln b = b^x \ln b.

Thus the derivative of an exponential function is

\frac{d} {dx} [b^x] = b^x \ln b.

In the special case where the base is b^x = e^x, since \ln e = 1 the derivative rule becomes

\frac{d} {dx} [e^x] = e^x.

To generalize, if u is a differentiable function of x, with the use of the Chain Rule the above derivatives take the general form

\frac{d} {dx} [b^u] = b^u \cdot \ln b \cdot \frac{du} {dx}.

And if b = e,

\frac{d} {dx} [e^u] = e^u \cdot \frac{du} {dx}.

Derivatives of Exponential Functions

\frac{d} {dx} [b^u] & = b^u \cdot \ln b \cdot \frac{du} {dx} = u' b^u \ln b\\\frac{d} {dx} [e^u] & = e^u \cdot \frac{du} {dx} = u' e^u

Example 9:

Find the derivative of y = 2^{x^2}.

Solution:

Applying the rule for differentiating an exponential function,

y' & = (2x) 2^{x^2} \ln 2\\& = 2^{x^2+ 1} \cdot x \cdot \ln 2.

Example 10:

Find the derivative of  y = e^{x^2}.

Solution:

Since

\frac{d} {dx} [e^u] & = u' e^u,\\y' & = 2x e^{x^2}.

Example 11:

Find f'(x) if

f(x) = \frac{1} {\sqrt{\pi \sigma}} e^{-\alpha k(x - x_0)^2}.

where \sigma, \alpha, x_0 , and k are constants and \sigma \neq 0.

Solution:

We apply the exponential derivative and the Chain Rule:

f'(x) & = \frac{1} {\sqrt{\pi \sigma}} (-2 \alpha k(x - x_0)) e^{-\alpha k(x - x_0)^2}\\& = - \frac{2 \alpha k (x - x_0)} {\sqrt{\pi \sigma}} e^{- \alpha k (x - x_0)^2}.

Integrals Involving Exponential Functions

Associated with the exponential derivatives in the box above are the two corresponding integration formulas:

\int b^u du & = \frac{1} {\ln b} b^u + C,\\\int e^u du & = e^u + C.

The following examples illustrate how they can be used.

Example 12:

Evaluate  \int 5^x dx.

Solution:

\int 5^x dx & = \frac{1} {\ln 5} 5^x + C\\& = \frac{5^x} {\ln 5} + C.

Example 13:

\int e^x dx.

Solution:

\int e^x dx = e^x + C.

In the next chapter, we will learn how to integrate more complicated integrals, such as  \int x^2 e^{x^3}dx, with the use of u-substitution and integration by parts along with the logarithmic and exponential integration formulas.

Multimedia Links

For a video presentation of the derivatives of exponential and logarithmic functions (4.4), see Math Video Tutorials by James Sousa, The Derivatives of Exponential and Logarithmic Functions (8:26).

Review Questions

  1. Find dy/dx of y = e^{6x}
  2. Find  dy/dx of  y = e^{3x^3 - 2x^2 + 6}
  3. Find  dy/dx of  y = e^{x^2} \cdot \ln \left (\frac{1} {x}\right )
  4. Find  dy/dx of  y = \frac{e^x - e^{-x}} {e^x + e^{-x}}
  5. Find dy/dx of y = \cos(e^x)
  6. Find dy/dx of y = \ln(\sin(\ln x))
  7. Evaluate  \int \frac{1} {e^x} dx
  8. Evaluate  \int \sqrt{e^x} dx
  9. Evaluate  \int \frac{4x - 3} {4x^2 - 6x + 7} dx
  10. Evaluate  \int \frac{e^x + e^{-x}} {e^x - e^{-x}} dx
  11. Evaluate  \int_0^e \frac{dx} {x + e}
  12. Evaluate  \int_{-\ln 3}^{\ln 3} \frac{e^x} {e^x + 4} dx

Review Answers

  1. y' = 6e^{6x}
  2.  y' = (9x^2 - 4x)e^{3x^3 - 2x^2 + 6}
  3.  y' = -e^{x^2} [2x \ln x + \frac{1}{x}]
  4.  y' = \frac{4} {(e^x + e^{-x})^2}
  5. y' = -e^x \cot e^x
  6.  y' = \frac{\cot(\ln x)} {x}
  7. -e^{-x} + C
  8. 2e^{x/2} + C
  9.  \frac{1} {2} \ln |4x^2 - 6x + 7| + C
  10.  \ln |e^x - e^{-x}| + C
  11. \ln 2
  12.  \ln \frac{21} {13}

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Feb 23, 2012

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Aug 21, 2014
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CK.MAT.ENG.SE.1.Calculus.6.3

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