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Derivatives of \(b^{x}\) Find the derivatives of the following functions. $$y=x^{3} \cdot 3^{x}$$

Short Answer

Expert verified
Answer: The derivative of the function is \(y'=3^x x^2 (3 + x\ln 3)\).

Step by step solution

01

Identify the Functions

First, let's identify the polynomial function and the exponential function in the given function. The polynomial function \(u(x)=x^3\) and the exponential function \(v(x)=3^x\).
02

Apply the Product Rule

The product rule states that the derivative of the product of two functions is the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function, i.e., $$(uv)' = u'v + uv'.$$ Applying the product rule to our function \(y=u(x)v(x)\), we get: $y' = u'(x)v(x) + u(x)v'(x).$$ Now, we need to find the derivatives \(u'(x)\) and \(v'(x)\).
03

Differentiate the Polynomial Function

The polynomial function is \(u(x)=x^3\). To find its derivative, we apply the power rule, which states \((x^n)'=nx^{n-1}\). Therefore, $$u'(x) = \frac{d}{dx}(x^3) = 3x^2.$$
04

Differentiate the Exponential Function

The exponential function is \(v(x)=3^x\). To find its derivative, we apply the chain rule and the rule for the derivative of \(b^x\), which is \((b^x)'=b^x\ln b\). Therefore, $$v'(x) = \frac{d}{dx}(3^x) = 3^x\ln 3.$$
05

Calculate the Derivative of the Given Function

Now that we have the derivatives \(u'(x)\) and \(v'(x)\), we can substitute them into the expression for the derivative of the product: $$y' = u'(x)v(x) + u(x)v'(x) = (3x^2)(3^x) + (x^3)(3^x\ln 3).$$ Finally, we can factor out common terms: $$y' = 3^x x^2 (3 + x\ln 3).$$ The derivative of the given function is $$y'=3^x x^2 (3 + x\ln 3).$$

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Key Concepts

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

Product Rule
The product rule is a fundamental technique in calculus for finding the derivative of a product of two functions. It comes into play when you have an equation where one function is multiplied by another. In simple terms, the product rule tells you that the derivative of the product is not just the product of the derivatives. To apply the product rule, follow these steps:

  • Identify the two functions that make up the product.
  • Find the derivative of the first function.
  • Find the derivative of the second function.
  • Multiply the derivative of the first function by the second function.
  • Multiply the first function by the derivative of the second function.
  • Add the two results together to get the final derivative.

Writing it mathematically, for two functions, say, \( u(x) \) and \( v(x) \), the derivative, noted as \( (uv)' \), is given by:
\[ (uv)' = u'v + uv' \]The exercise you're working with utilizes the product rule because the given function is the product of a polynomial function \( x^3 \) and an exponential function \( 3^x \).
Exponential Function Derivative
When it comes to the derivatives of exponential functions, it's essential to understand that these derivatives don't behave like those of polynomials. An exponential function has the form \( b^x \), where \( b \) is a constant. To differentiate an exponential function, you apply the rule \( (b^x)' = b^x\ln(b) \).

In this particular exercise, we're working with the function \( 3^x \). Its derivative is derived by multiplying the function itself by the natural logarithm of its base, \( \ln(3) \). This derivative is given by:
\[ \frac{d}{dx}(3^x) = 3^x\ln(3) \]
The concept is critical because it allows us to handle functions where the variable is in the exponent, which is a common scenario in various fields such as finance, physics, and biology.
Chain Rule
The chain rule is a powerful tool in calculus used to find the derivative of a composite function. In other words, when you have a function within a function, as often occurs in exponential functions like \( b^{g(x)} \), the chain rule can help you find the derivative efficiently.

The chain rule states that to differentiate a composite function, you must take the derivative of the outer function and multiply it by the derivative of the inner function. When you have an exponential function with a variable base, such as \( x^3 \), that's part of a product, you'd not apply the chain rule directly, but the principles are fundamentally connected to understanding how derivatives work in composite situations.
The exercise involves an exponential function but does not directly require the use of the chain rule. However, understanding the chain rule is crucial for grasping the interplay of functions and their derivatives in more complex scenarios.
Power Rule
The power rule is a basic yet extremely important rule in differentiation, especially when dealing with polynomial functions like \( x^n \). The rule is straightforward: for any real number \( n \), the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \). So for a term like \( x^3 \), applying the power rule gives us:
\[ \frac{d}{dx}(x^3) = 3x^2 \]
This rule simplifies the process of differentiation by reducing a seemingly complex function down to a simpler one. In our exercise, the power rule helps to determine the derivative of the polynomial part of the function. It's essential to understand the power rule and to remember that it applies only to terms where the variable is raised to a constant power.

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Most popular questions from this chapter

Economists use production functions to describe how the output of a system varies with respect to another variable such as labor or capital. For example, the production function \(P(L)=200 L+10 L^{2}-L^{3}\) gives the output of a system as a function of the number of laborers \(L\). The average product \(A(L)\) is the average output per laborer when \(L\) laborers are working; that is \(A(L)=P(L) / L\). The marginal product \(M(L)\) is the approximate change in output when one additional laborer is added to \(L\) laborers; that is, \(M(L)=\frac{d P}{d L}\). a. For the given production function, compute and graph \(P, A,\) and \(M\). b. Suppose the peak of the average product curve occurs at \(L=L_{0},\) so that \(A^{\prime}\left(L_{0}\right)=0 .\) Show that for a general production function, \(M\left(L_{0}\right)=A\left(L_{0}\right)\).

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

A store manager estimates that the demand for an energy drink decreases with increasing price according to the function \(d(p)=\frac{100}{p^{2}+1},\) which means that at price \(p\) (in dollars), \(d(p)\) units can be sold. The revenue generated at price \(p\) is \(R(p)=p \cdot d(p)\) (price multiplied by number of units). a. Find and graph the revenue function. b. Find and graph the marginal revenue \(R^{\prime}(p)\). c. From the graphs of \(R\) and \(R^{\prime}\), estimate the price that should be charged to maximize the revenue.

Gone fishing When a reservoir is created by a new dam, 50 fish are introduced into the reservoir, which has an estimated carrying capacity of 8000 fish. A logistic model of the fish population is \(P(t)=\frac{400,000}{50+7950 e^{-0.5 t}},\) where \(t\) is measured in years.c. How fast (in fish per year) is the population growing at \(t=0 ?\) At \(t=5 ?\) d. Graph \(P^{\prime}\) and use the graph to estimate the year in which the population is growing fastest.

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \left(\sec ^{4} x \tan ^{2} x\right)$$

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