/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Find the derivative. \(3^{x} \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivative. \(3^{x} \cdot 5^{x}\)

Short Answer

Expert verified
The derivative of \(3^x \cdot 5^x\) is \(15^x \ln(15)\).

Step by step solution

01

Rewrite the Expression using Exponent Properties

The expression given is \(3^x \cdot 5^x\). We can rewrite this as \((3 \cdot 5)^x\) because both terms have the same exponent \(x\). Simplifying further, we have \(15^x\).
02

Use the Power Rule for Exponential Functions

The derivative of \(a^x\) with respect to \(x\) is given by \(a^x \ln(a)\). Here, \(a = 15\). We apply the power rule for derivatives to differentiate \(15^x\).
03

Write the Derivative

The derivative of \(15^x\) with respect to \(x\) is \(15^x \ln(15)\). Therefore, \(\frac{d}{dx}(3^x \cdot 5^x) = 15^x \ln(15)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. A common form is \(a^x\), where \(a\) is the base and \(x\) is the exponent. These functions are unique because the variable is in the exponent, contrasting with polynomial functions where the variable is the base. Exponential growth and decay are key applications, frequently appearing in sciences. For example:
  • Population growth models
  • Radioactive decay
  • Compound interest calculations
Understanding exponential functions is crucial, as they help describe situations where the change is multiplicative rather than additive.
In the context of derivatives, these functions provide interesting challenges, such as determining the rate of growth or decay represented by the function. Recognizing a function as exponential, like \(3^x\), allows us to apply specific rules for differentiation, ensuring we capture the unique rate of change characteristics.
Power Rule
The power rule is a fundamental technique used in calculus to find the derivative of polynomial functions. It states that if you have a function \(x^n\), the derivative is \(nx^{n-1}\). This rule is straightforward and relies on decreasing the exponent by one and multiplying by the original exponent.
However, in the case of exponential functions, the power rule is slightly modified. When dealing with a function like \(a^x\) where \(a\) is a constant, the derivative is not simply handled by the power rule. Instead, it involves the natural logarithm of the base. The formula becomes \(a^x \ln(a)\).
This modification reveals how the base of the exponential governs its rate of change. As seen with \(15^x\), taking the derivative using the modified power rule prompts us to multiply by the logarithm of 15, resulting in \(15^x \ln(15)\) as the derivative.
Differentiation Techniques
Differentiation involves finding the rate at which a quantity changes. It's a crucial concept in calculus with methods tailored for various types of functions. Different techniques allow us to tackle a wide range of function types:
  • Exponential functions: For expressions like \(a^x\), we apply the exponential derivative formula \(a^x \ln(a)\).
  • Trigonometric functions: Utilize specific rules for sine, cosine, etc.
  • Product rule: Used when dealing with the multiplication of two functions.
  • Chain rule: Applied when differentiating composite functions.
In our example, combining \(3^x\cdot5^x\) initially may suggest using the product rule. However, recognizing both share the same exponent allows simplification to \(15^x\), showing the importance of understanding properties behind expressions. Differentiation is not just about applying rules, but also simplifying effectively to make problems manageable. The strategy of rewriting terms leverages inherent simplifications, showcasing calculus as both an art of transformation and a science of calculation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Biology Talekar and coworkers \(^{91}\) showed that the number \(y\) of the eggs of \(O\). furnacalis per mung bean plant was approximated by \(y=f(x)=-57.40+63.77 x-\) \(20.83 x^{2}+2.36 x^{3},\) where \(x\) is the age of the mung bean plant and \(2 \leq x \leq 7\). Find \(f^{\prime}(x),\) and explain what this means. Use the quadratic formula to show that \(f^{\prime}(x)>0\) for \(2 \leq x \leq 7\). What does this say about the preference of O. furnacalis to laying eggs on older plants?

Marginal Cost, Revenue, Profit In \(1997,\) Fuller and coworkers \({ }^{14}\) at Texas \(\mathrm{A} \& \mathrm{M}\) University estimated the operating costs of cotton gin plants of various sizes. A power model of cost in thousands of dollars for the next-to-thelargest plant is given by \(C(x)=165.46 x^{-0.4789},\) where \(x\) is the annual quantity of bales in thousands produced if annual production is at most 30,000 bales. Revenue was estimated at \(\$ 63.25\) per bale. a. Find the marginal cost. b. Find the marginal revenue. c. Find the marginal profit.

Biology Theoretical studies of photosynthesis \({ }^{31}\) assume that the gross rate \(R\) of photosynthesis per unit of leaf area is given by $$ R(E)=\frac{a E}{1+b E} $$ where \(E\) is the incident radiation per unit leaf area, and \(a\) and \(b\) are positive constants. Find the rate of change of \(R\) with respect to \(E\).

Economies of Scale In 1955 Surdis \(^{87}\) obtained records from a utility company regarding its trench digging operations. The records show that the unit cost \(C(n)\) per foot of earth removed by the mechanical trencher is given approximately by $$ C(n)=\frac{15.04+0.74 n}{25 n} $$ where \(n\) is the number of hours worked per day. a. Graph. Find values for \(C^{\prime}(n)\) at \(x=2,4,6,\) and \(8 .\) Interpret what these numbers mean. What is happening? Units costs for hand digging was found to be \(\$ 0.60 .\) b. Approximate the number of hours worked at which using the trench digging machinery is more cost effective than hand digging.

Take \(x>0\) and find the derivatives. \(\sqrt{\ln x}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.