/*! 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 10 Find the derivatives of the func... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the derivatives of the functions. Assume that \(a\) and \(k\) are constants. $$y=2^{x}+\frac{2}{x^{3}}$$

Short Answer

Expert verified
The derivative is \(\frac{dy}{dx} = 2^x \ln 2 - 6x^{-4}\).

Step by step solution

01

Identify the Function Types

The function given, \(y=2^x+\frac{2}{x^3}\), is made up of an exponential function \(2^x\) and a rational function \(\frac{2}{x^3}\). We need to find the derivative of each part separately.
02

Find Derivative of Exponential Function

To find the derivative of \(2^x\), use the rule \(\frac{d}{dx}a^x = a^x \ln a\). For \(2^x\), this gives: \(\frac{d}{dx}2^x = 2^x \ln 2\).
03

Apply Power Rule to Rational Term

Rewrite \(\frac{2}{x^3}\) as \(2x^{-3}\) and apply the power rule \(\frac{d}{dx}x^n = n x^{n-1}\). This gives \(\frac{d}{dx}2x^{-3} = 2(-3)x^{-4} = -6x^{-4}\).
04

Combine Derivatives

Combine the derivatives of the individual terms to find the derivative of the entire function. Thus, \(\frac{dy}{dx} = 2^x \ln 2 - 6x^{-4}\).

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Key Concepts

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

Exponential Function
An exponential function is any function where a constant base is raised to a variable exponent. In the given problem, the function is represented as \(2^x\). Here's why it's called exponential:
  • Constant Base: The base number here is 2, and it remains constant.
  • Variable Exponent: The exponent is the variable \(x\), allowing the function to grow rapidly as \(x\) increases.
To find the derivative of an exponential function like \(a^x\), where \(a\) is a constant, the rule is: \(\frac{d}{dx}a^x = a^x \ln a\). This means you not only multiply the original function by the natural logarithm of the base \(a\).
For our function, \(\frac{d}{dx}2^x = 2^x \ln 2\). This derivative tells us the rate of change of the exponential function at any given \(x\) value. Exponential functions increase or decrease very quickly, depending on whether \(a\) is more than 1 or between 0 and 1. Understanding how to differentiate these is crucial because they model phenomena like population growth and radioactive decay.
Power Rule
Differentiating terms that look like \(x^n\) is where the power rule comes into play. The power rule is a central tool in calculus used to find derivatives of polynomial functions, including those with negative exponents.
To apply the power rule, follow this formula: \(\frac{d}{dx}x^n = n x^{n-1}\). This involves multiplying the original exponent \(n\) by the coefficient in front of \(x^n\) and decreasing the exponent by one.
In our problem, we first rewrite the rational term \(\frac{2}{x^3}\) as \(2x^{-3}\) to apply the power rule more easily. The derivative then becomes \(-6x^{-4}\) because:
  • The original exponent \(-3\) gets multiplied by the coefficient 2, resulting in \(-6\).
  • The new exponent is \(-3 - 1 = -4\).
This simplified approach makes it easier to handle derivatives of terms presented as quotients or fractions. The power rule is an efficient method to differentiate any polynomial term.
Rational Function
A rational function is typically expressed as the quotient of two polynomials \(\frac{P(x)}{Q(x)}\). In the problem, this is shown by \(\frac{2}{x^3}\), where the numerator is 2 (a constant and itself a polynomial of degree 0) and the denominator is \(x^3\).
Rational functions can often be rewritten to make differentiation easier. Here, we convert \(\frac{2}{x^3}\) into \(2x^{-3}\), which behaves similarly to polynomial terms for which the power rule can be applied.
  • Expressing it as \(2x^{-3}\) allows direct application of the power rule, which simplifies finding the derivative.
  • The negative exponent also indicates the function decreases in value as \(x\) becomes larger.
Rational functions can be more complex when the numerator or the denominator becomes a higher degree polynomial, but the same principles apply. Understanding how to convert and manage these is vital to solving real-world scenarios involving rates of change and limits.

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Most popular questions from this chapter

Give an example of: A curve that has two horizontal tangents at the same \(x\) -value, but no vertical tangents.

True or false for a function \(f\) whose domain is all real numbers? If a statement is true, explain how you know. If a statement is false, give a counterexample. If \(f^{\prime}(x) \geq 0\) for all \(x,\) then \(f(a) \leq f(b)\) whenever \(a \leq b\).

The following problem is a generalization of Problem \(31 .\) Show that any function of the form $$y=A e^{x}+B e^{-x}, \quad A>0, B>0$$ can be written, for some \(K\) and \(c,\) in the form $$y=K \cosh (x-c)$$ What does this tell you about the graph of \(y=A e^{x}+\) \(B e^{-x} ?\)

A rubber duck bounces up and down in a pool after a stone is dropped into the water. The height of the duck, in inches, above the equilibrium position of the water is given as a function of time \(t,\) in seconds, by $$d(t)=e^{-t}(\cos t+\sin t)$$ (a) Find and interpret the practical meaning of the derivative \(d^{\prime}(t)\) (b) Determine when \(d^{\prime}(t)=0\) for \(t \geq 0 .\) What can you say about the duck when \(d^{\prime}(t)=0 ?\) (c) Determine \(\lim _{t \rightarrow \infty} d(t)\) and explain why this limit makes sense in practical terms.

With \(t\) in years, the population of a herd of deer is represented by $$P(t)=4000+500 \sin \left(2 \pi t-\frac{\pi}{2}\right)$$ (a) How does this population vary with time? Graph \(P(t)\) for one year. (b) When in the year the population is a maximum? What is that maximum? Is there a minimum? If so, when? (c) When is the population growing fastest? When is it decreasing fastest? (d) How fast is the population changing on July \(1 ?\)

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