Problem 31 Find $f^{\prime}(x)$ $$ f(... [FREE SOLUTION]

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Calculus Applications and Technology

Edmond C. Tomastik

$Math Studyset 91影视 Explanations$ Math

3 Edition

Chapter 4: Problem 31

Find $f^{\prime}(x)$ $$ f(x)=e^{4}-\ln x-0.01 x^{2}-0.03 x^{3} $$

Short Answer

Expert verified

$f'(x) = -\frac{1}{x} - 0.02x - 0.09x^2$.

Step by step solution

Identify the Derivative Rules

To differentiate the function $f(x) = e^4 - \ln x - 0.01x^2 - 0.03x^3$, we need to use the power rule for the last two terms and the derivative of natural logarithm and constants. The derivative of $e^4$, a constant, is 0. The derivative of $\ln x$ is $\frac{1}{x}$. For $-0.01x^2$ and $-0.03x^3$, we use the power rule where $d/dx\, x^n = nx^{n-1}$.

Differentiate Each Term

For the function given, differentiate each term as follows:- The derivative of the constant $e^4$ is 0.- The derivative of $-\ln x$ is $-\frac{1}{x}$.- The derivative of $-0.01x^2$ is $-0.01 \cdot 2x = -0.02x$.- The derivative of $-0.03x^3$ is $-0.03 \cdot 3x^2 = -0.09x^2$.

Combine the Derivatives

Add the derivatives of all terms together to find the derivative of the entire function: \[ f'(x) = 0 - \frac{1}{x} - 0.02x - 0.09x^2 \].

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Key Concepts

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

Derivative Rules

Differentiation is a key concept in calculus that involves determining the rate at which a function is changing at any given point. When tackling this, a few derivative rules become very handy:

Constant Rule: The derivative of a constant is always zero. This applies to numbers without variables, like $ e^4 $ in our exercise.
Sum Rule: If you have a sum of multiple functions, the derivative of this sum is the sum of the derivatives. This helps simplify problems with multiple terms.
Difference Rule: Similar to the sum rule, but it applies to subtraction. The derivative of a difference is the difference of the derivatives.

Keep these rules in mind as they can make differentiation much more approachable and manageable. They were crucial in solving our example problem.

Power Rule

The power rule is one of the simplest and most useful tools in differentiation. It is used whenever you have a term of the form $x^n$, where $n$ is any real number.
The rule states:

To differentiate $x^n$, multiply by the exponent $n$ and subtract one from the exponent: $\frac{d}{dx} x^n = nx^{n-1}$.

In the original exercise, the power rule helps us find the derivatives of $-0.01x^2$ and $-0.03x^3$.
Here's how it works:

For $-0.01x^2$, apply the power rule: Multiply $-0.01$ by 2 and reduce the exponent by one: the result is $-0.02x$.
For $-0.03x^3$, multiply $-0.03$ by 3, then subtract one from the exponent: the answer is $-0.09x^2$.

The power rule simplifies the differentiation process considerably, especially with polynomial expressions.

Natural Logarithm Derivative

The derivative of the natural logarithm function, $ \ln x $, is a fundamental aspect of calculus. When you differentiate $ \ln x $, the result is straightforward:

The derivative of $ \ln x $ is $ \frac{1}{x} $.

This comes handy in situations where logarithms are involved, such as the problem we dissected.
In the example, the function $ f(x) = e^4 - \ln x - 0.01x^2 - 0.03x^3 $ includes the term $-\ln x$. Differentiating $-\ln x$ using the rule above gives:

$-\ln x$ differentiates to $-\frac{1}{x}$.

Understanding how logarithms differentiate is crucial, particularly in applications involving growth processes or decay scenarios where natural logarithms frequently appear.

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91影视

Find \(f^{\prime}(x)\) $$ f(x)=e^{4}-\ln x-0.01 x^{2}-0.03 x^{3} $$

Short Answer

Step by step solution

Identify the Derivative Rules

Differentiate Each Term

Combine the Derivatives

Key Concepts

Derivative Rules

Power Rule

Natural Logarithm Derivative

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