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Chapter 1: Problem 84

Use a derivative routine to obtain the value of the derivative. Give the value to 5 decimal places. $$f^{\prime}(0), \text { where } f(x)=10^{1+x}$$

Short Answer

Expert verified
\[f^{\prime}(0) = 23.02585\].

Step by step solution

01

Understand the Derivative

The task is to find the derivative of the function at a specified point. Here, we need to find the derivative of the function at x = 0 of the function \[f(x) = 10^{1+x}\].
02

Write Down the Function

The given function is \[f(x) = 10^{1+x}\]. We need to find \[f^{\prime}(0)\].
03

Use the Chain Rule

To find the derivative of \[10^{1+x}\], we apply the chain rule. Let \[u = 1+x\]. Then \[f(x) = 10^{u}\]. The derivative is given by \[ f'(x) = 10^{u} \times \frac{d}{dx}(u) \times \frac{d}{du}(10^{u})\].
04

Evaluate the Derivative

First, compute \[\frac{d}{du}(10^{u}) = 10^{u} \times \text{ln}(10)\]. Now, \[u = 1+x\], so \[\frac{d}{dx}(u) = 1\]. Therefore, \[f'(x) = 10^{1+x} \times \text{ln}(10) \].
05

Substitute x = 0

Substitute \[x = 0\] into the derivative to find \[f^{\prime}(0)\]: \[f^{\prime}(0) = 10^{1+0} \times \text{ln}(10) = 10 \times \text{ln}(10)\].
06

Calculate the Value

Calculate \[10 \times \text{ln}(10)\] to 5 decimal places. The natural logarithm of 10 is approximately 2.302585. Thus, \[10 \times 2.302585 = 23.02585\].

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Key Concepts

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

Chain Rule
The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. A composite function is when one function is nested inside another. For example, in the function \(f(x) = 10^{1+x}\), you can think of \(1+x\) as a function \(u\) and \(10^u\) as another function. The chain rule allows us to differentiate these complex functions by differentiating the outer function and then the inner function. In our case:
  • First, we set \(u = 1 + x\).
  • Then we differentiate \(f(u) = 10^u\) with respect to \(u\).
  • We also need to differentiate \(u = 1 + x\) with respect to \(x\).
The chain rule formula is given by: \[ f'(x) = f'(u) \times \frac{du}{dx} \] Here, \(f'(u)\) is the derivative of \(10^u\) and \(\frac{du}{dx}\) is the derivative of \(1+x\) with respect to \(x\).
Natural Logarithm
The natural logarithm, denoted as \(\ln\), is the logarithm to the base e, where e is approximately equal to 2.71828. When differentiating exponential functions, natural logarithms often come into play.
In our exercise, we need to differentiate \(10^u\). To do this, we use the fact that the derivative of \(a^x\) (where a is a constant) is \(a^x \ln(a)\). For example:
  • To differentiate \(10^u\) with respect to \(u\), we have \(\frac{d}{du}(10^u) = 10^u \ln(10)\).
This rule comes from the general property of logarithms and exponentials. When we apply this to our function, it simplifies the differentiation process by converting the exponential term into a product of the term itself and the natural logarithm.
Differentiation
Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which the function's value changes as its input changes. In basic terms, it's a measure of how a function is changing at any given point.
Here鈥檚 how differentiation was applied in the exercise:
  • We started with \(f(x) = 10^{1+x}\).
  • We used the chain rule: Let \(u = 1 + x\).
  • Thus, the function transforms to \(10^u\).
  • Using the rule for differentiating \(a^x\), the derivative of \(10^u\) with respect to \(u\) is \(10^u \ln(10)\).
  • The derivative of \(u = 1 + x\) with respect to \(x\) is 1.
Hence, the derivative of the original function \(f(x)\) with respect to \(x\) is \(10^{1+x} \ln(10)\).
We then evaluated this at \(x = 0\), giving us \[f'(0) = 10^{1+0} \ln(10) = 10 \ln(10)\]. Finally, we calculated this value to 5 decimal places, obtaining approximately 23.02585.

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Most popular questions from this chapter

If \(f(x)\) is a linear function, \(f(1)=0,\) and \(f(2)=1,\) what is \(f(3) ?\)

Find the indicated derivative. $$\frac{d}{d x}\left(x^{-3}\right)$$

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