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Chapter 4: Problem 39

Differentiate the functions with respect to the independent variable. \(f(x)=2^{x^{2}+1}\)

Short Answer

Expert verified
The derivative is \( \frac{df}{dx} = 2^{x^2 + 1} \cdot 2x \cdot \ln(2) \).

Step by step solution

01

Identify the function

The given function is \( f(x) = 2^{x^2 + 1} \). This is an exponential function where the base is a constant (2) and the exponent is a function of \( x \).
02

Apply the Chain Rule

To differentiate \( f(x) = 2^{x^2 + 1} \), use the chain rule. The derivative of \( a^u \), where \( a \) is a constant and \( u \) is a function, is \( a^u \ln(a) \cdot \frac{du}{dx} \). Here, \( a = 2 \) and \( u = x^2 + 1 \).
03

Differentiate the exponent

Differentiate the exponent \( u = x^2 + 1 \) with respect to \( x \). The derivative \( \frac{du}{dx} = \frac{d}{dx}(x^2 + 1) = 2x \).
04

Combine derivatives

Combine the results from Steps 2 and 3. The derivative of \( f(x) = 2^{x^2 + 1} \) is given by:\[ \frac{df}{dx} = 2^{x^2 + 1} \cdot \ln(2) \cdot 2x \]Simplify this to:\[ \frac{df}{dx} = 2^{x^2 + 1} \cdot 2x \cdot \ln(2) \]

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Key Concepts

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

Exponential Function Differentiation
In calculus, when we talk about exponential function differentiation, we're usually referring to functions where the base is a constant and the exponent involves a variable. An example is the function \( f(x) = 2^{x^2+1} \). Here, 2 is a constant base, and the exponent is the expression \( x^2 + 1 \), which depends on \( x \).
Exponential functions are everywhere, since they represent rapid growth or decay, like population growth or radioactive decay.
  • The General Rule: If you have a function in the form \( a^u \), where \( a \) is a constant, its derivative involves multiplying the function by the natural logarithm of its base.
  • In our example, the derivative will start with \( 2^{x^2 + 1} \cdot \ln(2) \).
This step lays the foundation for what comes next, using the intricacies of the chain rule.
Chain Rule in Differentiation
The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. It becomes particularly crucial in scenarios involving exponential functions with variable exponents, like \( f(x) = 2^{x^2 + 1} \).
This rule helps us break down and understand how the rate of change inside the function affects the overall rate of the exponential function.
  • What does the Chain Rule say? It says that to differentiate a composition of functions: you first differentiate the outer function (with respect to its inner function) and multiply by the derivative of the inner function.
  • For \( f(x) \), apply the chain rule to obtain: \( 2^{x^2 + 1} \cdot \ln(2) \cdot \frac{du}{dx} \).
The magical part of the chain rule is how it connects inside and outside changes, letting us compute derivatives gracefully and effectively.
Derivative of an Exponent
Finding the derivative of an exponent involves differentiating the expression within the exponent. In \( f(x) = 2^{x^2 + 1} \), the exponent is the quadratic \( x^2 + 1 \).
We need to differentiate this expression to complete our chain rule process.
  • Differentiate the Exponent: If \( u = x^2 + 1 \), then \( \frac{du}{dx} = 2x \), because the derivative of \( x^2 \) is \( 2x \) and the constant 1 becomes zero upon differentiation.
  • Combine this with the chain rule: The full derivative of \( f(x) \) is \( 2^{x^2 + 1} \cdot \ln(2) \cdot 2x \).
Understanding this part reinforces the application of the chain rule and highlights the role exponentials play in differentiating functions efficiently and consistently.

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Most popular questions from this chapter

Suppose that the per capita growth rate of a population is \(3 \%\); that is, if \(N(t)\) denotes the population size at time \(t\), then $$\frac{1}{N} \frac{d N}{d t}=0.03$$ Suppose also that the population size at time \(t=4\) is equal to 100 . Use a linear approximation to compute the population size at time \(t=4.1\).

Find \(c\) so that \(f^{\prime}(c)=0 .\) \(\cos (\pi-x)\)

A measurement error in \(x\) affects the accuracy of the value \(f(x) .\) In each case, determine an interval of the form $$[f(x)-\Delta f, f(x)+\Delta f]$$ that reflects the measurement error \(\Delta x .\) In each problem, the quantities given are \(f(x)\) and \(x=\) true value of \(x \pm|\Delta x| .\) \(f(x)=1-3 x, x=-2 \pm 0.3\)

Find the derivatives of the following functions: $$ f(x)=\cos \sqrt{x^{2}+1} $$

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\frac{1}{1-x}\) at \(a=0\)
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