Derivatives of Exponential & Logarithmic Functions: Videos & Practice Problems

In calculus, understanding how to find the derivative of exponential functions is essential. The derivative of an exponential function of the form \( b^x \), where \( b \) is a positive constant not equal to 1, can be derived using the limit definition of the derivative. This process begins with the expression:

\[f'(x) = \lim_{h \to 0} \frac{b^{x+h} - b^x}{h}\]

By applying properties of exponents, this can be rewritten as:

\[f'(x) = \lim_{h \to 0} \frac{b^x \cdot b^h - b^x}{h} = b^x \cdot \lim_{h \to 0} \frac{b^h - 1}{h}\]

It turns out that the limit \( \lim_{h \to 0} \frac{b^h - 1}{h} \) is equal to \( \ln(b) \), the natural logarithm of \( b \). Therefore, the derivative of \( b^x \) is given by:

\[\frac{d}{dx}(b^x) = b^x \cdot \ln(b)\]

This rule is crucial for differentiating exponential functions, and it is important to remember that the base \( b \) must be greater than 0 and not equal to 1 to ensure the function is valid.

For example, to find the derivative of \( f(x) = 6^x \), we apply the rule:

\[f'(x) = 6^x \cdot \ln(6)\]

In cases where the exponent is a function of \( x \), such as \( g(x) = 3^{x^2 + 4x} \), we must also use the chain rule. The derivative is calculated as follows:

\[g'(x) = 3^{x^2 + 4x} \cdot \ln(3) \cdot \frac{d}{dx}(x^2 + 4x)\]

Calculating the derivative of the inner function \( x^2 + 4x \) gives \( 2x + 4 \). Thus, the full derivative becomes:

\[g'(x) = 3^{x^2 + 4x} \cdot \ln(3) \cdot (2x + 4)\]

By mastering these techniques, students can confidently differentiate a wide range of exponential functions, reinforcing their understanding of calculus principles.

In calculus, understanding how to differentiate exponential functions is crucial, particularly when dealing with the natural exponential function, \( e^x \). The derivative of \( e^x \) is unique because it is equal to the function itself. This can be derived from the general rule for exponential functions, which states that the derivative of \( b^x \) is \( b^x \ln(b) \). For \( e^x \), since the base \( b \) is \( e \), the derivative simplifies to:

\[\frac{d}{dx} e^x = e^x \cdot \ln(e) = e^x \cdot 1 = e^x\]

This property extends to functions multiplied by constants. For example, the derivative of \( 4e^x \) is simply \( 4 \) times the derivative of \( e^x \), which results in:

\[\frac{d}{dx} (4e^x) = 4e^x\]

When dealing with more complex functions involving \( e^x \), such as \( f(x) = 3e^{2x+4} \), the chain rule becomes necessary. The chain rule states that if you have a composite function, the derivative is the derivative of the outer function multiplied by the derivative of the inner function. For \( f(x) \), the derivative is calculated as follows:

\[f'(x) = 3 \cdot e^{2x+4} \cdot \frac{d}{dx}(2x+4) = 3 \cdot e^{2x+4} \cdot 2 = 6e^{2x+4}\]

Another example involves the product of a polynomial and an exponential function, such as \( g(x) = x e^{5x} \). Here, the product rule is applied, which states that the derivative of a product of two functions is given by:

\[\frac{d}{dx}(u \cdot v) = u'v + uv'\]

In this case, let \( u = x \) and \( v = e^{5x} \). The derivatives are \( u' = 1 \) and \( v' = e^{5x} \cdot 5 \) (using the chain rule). Thus, the derivative \( g'(x) \) is calculated as follows:

\[g'(x) = x \cdot (5e^{5x}) + e^{5x} \cdot 1 = 5xe^{5x} + e^{5x}\]

Factoring out \( e^{5x} \) gives:

\[g'(x) = e^{5x}(5x + 1)\]

These examples illustrate the application of the derivative rules for exponential functions, including the chain rule and product rule, which are essential for solving more complex calculus problems involving \( e^x \).

In calculus, understanding the derivatives of exponential functions is crucial, particularly for the natural exponential function, \( e^x \). The derivative of \( e^x \) can be derived from the general rule for exponential functions, which states that the derivative of \( b^x \) is given by:

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

For the specific case of \( e^x \), where the base \( b \) is \( e \), the derivative simplifies significantly because the natural logarithm of \( e \) is 1. Thus, the derivative of \( e^x \) is:

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

This property indicates that the function \( e^x \) is unique in that it is its own derivative. When dealing with functions that include a constant multiplier, such as \( 4e^x \), the constant can be factored out, leading to:

\[\frac{d}{dx} (4e^x) = 4 \cdot \frac{d}{dx} e^x = 4e^x\]

For more complex functions involving \( e^x \), such as \( f(x) = 3e^{2x+4} \), the chain rule must be applied. The chain rule states that if a function is composed of another function, the derivative is the derivative of the outer function multiplied by the derivative of the inner function. Thus, the derivative of \( f(x) \) is calculated as follows:

\[f'(x) = 3 \cdot e^{2x+4} \cdot \frac{d}{dx}(2x+4) = 3 \cdot e^{2x+4} \cdot 2 = 6e^{2x+4}\]

Another example involves the product of a polynomial and an exponential function, such as \( g(x) = x e^{5x} \). Here, the product rule is necessary, which states:

\[\frac{d}{dx}(u \cdot v) = u'v + uv'\]

Applying this to \( g(x) \), where \( u = x \) and \( v = e^{5x} \), we find:

\[g'(x) = \frac{d}{dx}(x) \cdot e^{5x} + x \cdot \frac{d}{dx}(e^{5x})\]

Calculating each derivative, we have:

\[g'(x) = 1 \cdot e^{5x} + x \cdot e^{5x} \cdot 5 = e^{5x} + 5xe^{5x}\]

Factoring out \( e^{5x} \) gives:

\[g'(x) = e^{5x}(5x + 1)\]

These examples illustrate the application of the derivative rules for exponential functions, including the chain rule and product rule, which are essential for solving more complex calculus problems involving \( e^x \).

To find the derivative of the function \( y = e^{\sqrt{x^3 - 2x}} \), we will utilize the chain rule, which is essential for differentiating composite functions. The chain rule states that if you have a function \( f(g(x)) \), the derivative is given by \( f'(g(x)) \cdot g'(x) \).

Starting with the outer function, which is \( e^u \) where \( u = \sqrt{x^3 - 2x} \), we know that the derivative of \( e^u \) is simply \( e^u \). Therefore, we have:

\( \frac{dy}{dx} = e^{\sqrt{x^3 - 2x}} \cdot \frac{d}{dx}(\sqrt{x^3 - 2x}) \)

Next, we need to differentiate the inner function \( \sqrt{x^3 - 2x} \). This can be rewritten as \( (x^3 - 2x)^{1/2} \). Applying the power rule, we bring down the exponent \( \frac{1}{2} \) and decrease the exponent by one:

\( \frac{d}{dx}(\sqrt{x^3 - 2x}) = \frac{1}{2}(x^3 - 2x)^{-1/2} \cdot \frac{d}{dx}(x^3 - 2x) \)

Now, we differentiate \( x^3 - 2x \) using the power rule:

\( \frac{d}{dx}(x^3 - 2x) = 3x^2 - 2 \)

Substituting this back into our expression gives:

\( \frac{d}{dx}(\sqrt{x^3 - 2x}) = \frac{1}{2}(x^3 - 2x)^{-1/2} \cdot (3x^2 - 2) \)

Now, we can combine everything to find the full derivative:

\( \frac{dy}{dx} = e^{\sqrt{x^3 - 2x}} \cdot \left( \frac{1}{2}(x^3 - 2x)^{-1/2} \cdot (3x^2 - 2) \right) \)

To simplify, we can express \( (x^3 - 2x)^{-1/2} \) as \( \frac{1}{\sqrt{x^3 - 2x}} \). Thus, the derivative can be rewritten as:

\( \frac{dy}{dx} = \frac{e^{\sqrt{x^3 - 2x}}(3x^2 - 2)}{2\sqrt{x^3 - 2x}} \)

This final expression represents the derivative of the function \( y \) with respect to \( x \), having applied the chain rule effectively. If you have any questions or need further clarification, feel free to ask!

To find the equation of the tangent line to the function \( y = 3e^x - x \) at the point where \( x = 0 \), we first need to determine the slope of the tangent line, which is given by the derivative of the function. The derivative, denoted as \( \frac{dy}{dx} \), can be calculated by differentiating each term of the function.

For the function \( y = 3e^x - x \), we apply the rules of differentiation. The derivative of \( 3e^x \) is \( 3e^x \), and the derivative of \( -x \) is \( -1 \). Therefore, the derivative of the function is:

\[ \frac{dy}{dx} = 3e^x - 1 \]

Next, we evaluate the derivative at \( x = 0 \) to find the slope of the tangent line at that point:

\[ \frac{dy}{dx} \bigg|_{x=0} = 3e^0 - 1 = 3 \cdot 1 - 1 = 2 \]

Now that we have the slope \( m = 2 \), we need to find the corresponding \( y \)-coordinate at \( x = 0 \) to use in the point-slope form of the equation of a line. We substitute \( x = 0 \) into the original function:

\[ y = 3e^0 - 0 = 3 \cdot 1 - 0 = 3 \]

Thus, the point of tangency is \( (0, 3) \). We can now use the point-slope form of the equation of a line, which is given by:

\[ y - y_1 = m(x - x_1) \]

Substituting \( m = 2 \), \( x_1 = 0 \), and \( y_1 = 3 \) into the equation, we have:

\[ y - 3 = 2(x - 0) \]

Rearranging this equation gives us:

\[ y - 3 = 2x \]

Finally, adding 3 to both sides results in the equation of the tangent line:

\[ y = 2x + 3 \]

This equation represents the tangent line to the function \( y = 3e^x - x \) at the point where \( x = 0 \).

To find the derivative of the logarithmic function \( y = \log_b(x) \), we can utilize the relationship between logarithms and exponentials. By rewriting the logarithmic equation in its exponential form, we have \( x = b^y \). This transformation allows us to apply the rules of differentiation to both sides of the equation.

When differentiating \( x = b^y \) with respect to \( x \), the derivative of \( x \) is simply 1. For the right side, we apply the chain rule, resulting in \( \frac{d}{dx}(b^y) = b^y \cdot \ln(b) \cdot \frac{dy}{dx} \). Setting these derivatives equal gives us the equation:

\( 1 = b^y \cdot \ln(b) \cdot \frac{dy}{dx} \)

To isolate \( \frac{dy}{dx} \), we rearrange the equation:

\( \frac{dy}{dx} = \frac{1}{b^y \cdot \ln(b)} \)

Since \( b^y = x \) (from our earlier transformation), we can substitute back to express the derivative in terms of \( x \):

\( \frac{dy}{dx} = \frac{1}{x \cdot \ln(b)} \)

This gives us the derivative of the logarithmic function \( \log_b(x) \) as:

\( \frac{d}{dx}(\log_b(x)) = \frac{1}{x \cdot \ln(b)} \)

It is important to note that the base \( b \) must be greater than 0 and not equal to 1, and \( x \) must be greater than 0, as logarithms are undefined for non-positive values.

For example, to find the derivative of \( g(x) = \log_8(x) \), we apply the rule:

\( g'(x) = \frac{1}{x \cdot \ln(8)} \)

Next, consider a more complex function, such as \( f(x) = \log_5(x^2) \). Here, we need to apply the chain rule. The derivative of \( \log_5(x^2) \) is:

\( f'(x) = \frac{1}{x^2 \cdot \ln(5)} \cdot \frac{d}{dx}(x^2) \)

Calculating the derivative of \( x^2 \) gives us \( 2x \), leading to:

\( f'(x) = \frac{2x}{x^2 \cdot \ln(5)} \)

After simplifying, we find:

\( f'(x) = \frac{2}{x \cdot \ln(5)} \)

These examples illustrate how to apply the derivative rules for logarithmic functions, including the use of the chain rule for more complex expressions.

To find the derivative of the function \( y = \log_4\left(\frac{\sqrt{x}}{4x + 1}\right) \), we will utilize several calculus rules, including the chain rule and the quotient rule. The derivative of a logarithmic function is given by the formula:

\[\frac{d}{dx} \log_b(u) = \frac{1}{u \ln(b)} \cdot \frac{du}{dx}\]

In our case, \( u = \frac{\sqrt{x}}{4x + 1} \) and \( b = 4 \). Thus, the derivative can be expressed as:

\[\frac{dy}{dx} = \frac{1}{\frac{\sqrt{x}}{4x + 1} \ln(4)} \cdot \frac{d}{dx}\left(\frac{\sqrt{x}}{4x + 1}\right)\]

Next, we need to differentiate \( u \) using the quotient rule, which states:

\[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}\]

Here, let \( f(x) = \sqrt{x} \) and \( g(x) = 4x + 1 \). The derivatives are:

\[f'(x) = \frac{1}{2\sqrt{x}} \quad \text{and} \quad g'(x) = 4\]

Applying the quotient rule gives us:

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

Now, simplifying the numerator:

\[(4x + 1)\left(\frac{1}{2\sqrt{x}}\right) - 4\sqrt{x} = \frac{4x + 1}{2\sqrt{x}} - 4\sqrt{x} = \frac{4x + 1 - 8x}{2\sqrt{x}} = \frac{1 - 4x}{2\sqrt{x}}\]

Thus, we have:

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

Substituting this back into our derivative expression, we get:

\[\frac{dy}{dx} = \frac{1}{\frac{\sqrt{x}}{4x + 1} \ln(4)} \cdot \frac{1 - 4x}{2\sqrt{x}(4x + 1)^2}\]

Now, simplifying further, we can multiply through by the reciprocal of \( \frac{\sqrt{x}}{4x + 1} \):

\[\frac{dy}{dx} = \frac{(1 - 4x)(4x + 1)}{2\sqrt{x}(4x + 1)^2 \ln(4) \cdot \sqrt{x}} = \frac{(1 - 4x)}{2\ln(4)(4x + 1)\sqrt{x}}\]

Finally, the derivative of the function \( y = \log_4\left(\frac{\sqrt{x}}{4x + 1}\right) \) is:

\[\frac{dy}{dx} = \frac{1 - 4x}{2\ln(4)(4x + 1)\sqrt{x}}\]

This result encapsulates the application of logarithmic differentiation, the chain rule, and the quotient rule, demonstrating the complexity of derivatives involving composite functions.

In calculus, understanding the derivative of logarithmic functions is essential, particularly the natural logarithm, denoted as ln(x) or log_e(x). The derivative of the natural logarithm can be derived from the general logarithmic function, where the base is e. The derivative of ln(x) is given by the formula:

Derivative of ln(x):
\[\frac{d}{dx} \ln(x) = \frac{1}{x} \quad \text{for } x > 0\]

This means that for any positive value of x, the slope of the tangent line to the curve of ln(x) is equal to the reciprocal of x.

When dealing with a constant multiplied by the natural logarithm, such as 6ln(x), the constant multiple rule applies. The derivative can be calculated as:

Example:
\[\frac{d}{dx} (6 \ln(x)) = 6 \cdot \frac{1}{x} = \frac{6}{x}\]

For more complex functions involving the natural logarithm, such as ln(f(x)), the chain rule must be applied. The chain rule states that if you have a composite function, the derivative is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function.

For instance, if we want to find the derivative of ln(x² + 4x), we start with the outer function, which is the natural logarithm:

Derivative using the chain rule:
\[\frac{d}{dx} \ln(x^2 + 4x) = \frac{1}{x^2 + 4x} \cdot (2x + 4) = \frac{2x + 4}{x^2 + 4x}\]

In another scenario, when finding the derivative of a product of functions, such as xln(x³), the product rule is utilized. The product rule states that the derivative of a product of two functions is:

Product Rule:
\[\frac{d}{dx}(u \cdot v) = u'v + uv'\]

Applying this to xln(x³), we first differentiate each part:

Example:
Let u = x and v = ln(x³). Then:

1. The derivative of u is u' = 1.
2. For v = ln(x³), we apply the chain rule:

\[\frac{d}{dx} \ln(x^3) = \frac{1}{x^3} \cdot 3x^2 = \frac{3}{x}\]

Now, applying the product rule:

\[g'(x) = 1 \cdot \ln(x^3) + x \cdot \frac{3}{x} = \ln(x^3) + 3\]

Thus, the final derivative is:

\[g'(x) = \ln(x^3) + 3\]

These examples illustrate the application of the derivative rules for logarithmic functions, including the natural logarithm, the chain rule, and the product rule, which are fundamental in calculus for analyzing the behavior of functions.

In this problem, we are given a function that describes the time t in days it takes for the number of bacteria in a sample to reach a certain amount b. The function is defined as:

\( t(b) = 42 \ln\left(\frac{650}{1500 - b}\right) \)

To analyze the growth of bacteria, we need to find the derivative of this function, denoted as \( t'(b) \), and evaluate it at \( b = 1000 \). The derivative represents the rate of change of time with respect to the number of bacteria, which can be interpreted as how the time required to reach a certain bacterial count changes as the count itself changes.

To find \( t'(b) \), we apply the chain rule and the properties of logarithmic differentiation. The derivative of the natural logarithm function is given by:

\( \frac{d}{dx} \ln(x) = \frac{1}{x} \)

Thus, we can express the derivative as:

\( t'(b) = 42 \cdot \frac{1}{1500 - b} \cdot \frac{d}{db}(1500 - b) \)

Calculating the derivative of \( 1500 - b \) gives us \(-1\), leading to:

\( t'(b) = \frac{42 \cdot (-1)}{1500 - b} = \frac{-42}{1500 - b} \)

Now, substituting \( b = 1000 \) into the derivative:

\( t'(1000) = \frac{-42}{1500 - 1000} = \frac{-42}{500} = -0.084 \)

This result indicates that at a bacterial count of 1,000, the time required to reach the next bacteria decreases at a rate of 0.084 days per additional bacterium. This negative value signifies that as the number of bacteria increases, the time taken to reach the next bacterium decreases, reflecting a faster growth rate.

To further interpret this, converting the rate into hours provides a clearer understanding. Since there are 24 hours in a day, we can express the growth rate as:

\( 0.084 \text{ days} \times 24 \text{ hours/day} \approx 2 \text{ hours} \)

This means that when the sample has 1,000 bacteria, it takes approximately 2 hours for the population to grow by one additional bacterium. Thus, the growth rate can be understood as the sample growing at a rate of one bacterium every 2 hours when the count is at 1,000.