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State the derivative rule for the exponential function \(f(x)=b^{x}\) How does it differ from the derivative formula for \(e^{x} ?\).

Short Answer

Expert verified
Answer: The main difference between the derivative rule for \(f(x) = b^x\) and the derivative formula for \(g(x) = e^x\) is that the derivative of an exponential function with base \(b\) has an extra factor of \(\ln{b}\) in its derivative formula compared to the derivative of the natural exponential function \(e^x\).

Step by step solution

01

Review the chain rule

The chain rule is a method used in calculus to differentiate composite functions. It states that the derivative of a composite function is the product of the derivative of the outer function times the derivative of the inner function. Mathematically, if \(y = f(g(x))\), then: $$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$ where \(u = g(x)\).
02

Derive the derivative rule for \(f(x) = b^x\)

To find the derivative of \(f(x) = b^x\), we will apply the chain rule. Let \(y = b^x\), then: $$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$ where \(u = x \ln{b}\). Now, we can find the derivatives: $$ \frac{dy}{du} = \frac{d(b^u)}{du} = b^u \ln{b} \\ \frac{du}{dx} = \frac{d(x \ln{b})}{dx} = \ln{b} $$ So, we can derive the derivative rule for \(f(x) = b^x\) as follows: $$ \frac{dy}{dx} = b^x \ln{b} $$
03

Derive the derivative rule for \(g(x) = e^x\)

To derive the derivative formula for \(g(x) = e^x\), we can use the fact that \(e\) is defined as: $$ e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n $$ By taking the derivative of \(g(x) = e^x\) with respect to \(x\), we find that: $$ \frac{dg}{dx} = e^x $$
04

Compare the derivative formulas

Now that we have derived the derivative formulas for both \(f(x) = b^x\) and \(g(x) = e^x\), we can compare them: - Derivative of \(f(x) = b^x\): $$ \frac{dy}{dx} = b^x \ln{b} $$ - Derivative of \(g(x) = e^x\): $$ \frac{dg}{dx} = e^x $$ The difference between the two is that the derivative of an exponential function with base \(b\) has an extra factor of \(\ln{b}\) in its derivative formula compared to the derivative of the natural exponential function \(e^x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
Understanding the chain rule is crucial for differentiating composite functions. It allows us to break down more complex expressions into simpler ones, which we can differentiate individually. For instance, when faced with a function like \(f(g(x))\), the derivative is not always straightforward. However, by applying the chain rule, which states \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), where \(u = g(x)\), we can tackle the outer function \(f\) and the inner function \(g\) separately.

Therefore, if one were to derive a formula for \(f(x) = b^x\) where \(b\) is a constant, using the chain rule makes this task manageable. The derivative of the outer function \(b^u\) with respect to \(u\) adds a factor of \(\ln{b}\) due to the properties of exponential functions, while the derivative of the inner function - in this case, simply \(x\) with respect to \(x\) - remains 1. This orchestrated process simplifies complex differentiation problems.
Composite Functions
Composite functions are functions made up of two or more simpler functions, where the output of one function becomes the input of another. Mathematically speaking, if we have two functions \(f\) and \(g\), their composition is written as \(f(g(x))\).

When it comes to differentiation, composite functions require special attention. This is where the chain rule plays a vital role, allowing us to differentiate a function of a function. The key to understanding composite functions is recognizing how the functions are intertwined and applying differentiation techniques to each part accordingly. By systematically applying the rules we know for derivatives, we can find the derivative of remarkably complex functions.
Natural Exponential Function
The natural exponential function, represented as \(e^x\), holds a special place in calculus. The number \(e\) is an irrational mathematical constant approximately equal to 2.71828, and it is the base of the natural logarithm. A defining property of \(e^x\) is that it is its own derivative, which means \(\frac{d}{dx}e^x = e^x\).

This unique property simplifies the process of differentiating functions involving \(e\) as the base and makes the natural exponential function a powerful tool in various calculus techniques, including solving differential equations. Its simplicity contrasts with other exponential functions, where additional steps are often required, such as multiplying by the natural logarithm of the base as seen in the derivative of \(b^x\).
Logarithmic Differentiation
Logarithmic differentiation is a method used in calculus to simplify the process of differentiating functions by applying the properties of logarithms. This technique is particularly useful when dealing with products, quotients, or powers of functions that are otherwise difficult to differentiate using standard rules.

By taking the natural logarithm of both sides of an equation and then differentiating, we can reach a more manageable form. For example, when differentiating \(b^x\), one could apply logarithmic differentiation by first expressing the function as \(y = b^x\) and then taking the natural logarithm to get \(\ln y = x \ln b\). Differentiating both sides with respect to \(x\) and applying the chain rule will lead us to \(\frac{dy}{dx} = b^x \ln{b}\), as seen in the solution for the derivative of \(b^x\). This approach simplifies the differentiation of functions that may not be immediately approachable by standard differentiation rules.

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Most popular questions from this chapter

A 500-liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min} .\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank (in liters) is given by \(V(t)=500-0.5 t\). a. Graph the mass function and verify that \(M(0)=0\). b. Graph the volume function and verify that the tank is empty when \(t=1000\) min. c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) and \(\lim _{\theta \rightarrow 000^{-}} C(t) ?\) \(t \rightarrow 1\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\). e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\). f. For what times is the concentration of the solution increasing? Decreasing?

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \frac{(2 x-1)(x+2)^{3}}{(1-4 x)^{2}}$$

Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. $$x^{4}=2\left(x^{2}-y^{2}\right) \text { (eight curve) }$$

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}(2 x)^{2 x}$$.

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$

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