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

Short Answer

Expert verified
The derivative of \( f(x) = \ln \sqrt{x^{2}+1} \) is \( \frac{x}{x^{2}+1} \).

Step by step solution

01

Simplify the Function

Simplify the function to make it easier to differentiate. Rewrite \( f(x) = \ln \sqrt{x^2 + 1} \) as \( f(x) = \ln (x^2 + 1)^{1/2} \). By using the logarithmic property, this can be expressed as \( \frac{1}{2} \ln(x^2 + 1) \).
02

Apply the Differentiation Rule

Differentiate the simplified function \( f(x) = \frac{1}{2}\ln(x^2 + 1) \). Use the chain rule and the derivative of the natural log function. The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).
03

Differentiate \( \ln(x^2 + 1) \)

Let \( u = x^2 + 1 \). Differentiate \( \ln(u) \) as \( \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{x^2+1} \cdot 2x \).
04

Combine the Results

Multiply the result from Step 3 by the constant \( \frac{1}{2} \) that was factored out earlier. Thus, the derivative \( f'(x) = \frac{1}{2} \cdot \frac{2x}{x^2+1} = \frac{x}{x^2+1} \).

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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 tool in calculus differentiation, particularly useful when dealing with composite functions. Think of a composite function as one function inside another, like wrapping an onion with different layers. To differentiate such functions, you'll need to understand how each layer interacts.

To apply the chain rule, differentiate the outer function first, while keeping the inner function unchanged. Then, multiply the result by the derivative of the inner function. This rule allows each component of a composite function to be tackled step-by-step. For example, in differentiating \( f(x) = \frac{1}{2}\ln(x^2 + 1) \), treat \( \ln(x^2 + 1) \) as a composite function where the outer function is \( \ln(u) \) and the inner function is \( u = x^2 + 1 \). Hence, apply the chain rule using the derivative of \( \ln(u) \) and the derivative of \( u \) itself.

By following the chain rule, you can decompose and differentiate complex functions with ease.
Logarithmic Differentiation
Logarithmic differentiation is a powerful technique used for differentiating functions where variables are raised to variable powers, or products and quotients of functions are involved. This method simplifies the differentiation process significantly.

The essence of logarithmic differentiation lies in taking the natural logarithm of both sides of the equation before differentiating. This clever technique makes tedious algebraic manipulations simpler, especially when dealing with powers, roots, and products.

Let's apply it to simplify \( f(x) = \ln \sqrt{x^2 + 1} \) into a form that's easier to work with. By rewriting it as \( f(x) = \frac{1}{2}\ln(x^2 + 1) \), you can differentiate with respect to \( x \) using the chain rule and straightforward derivative of the log function. The resulting process streamlines the computation and opens doors to differentiating otherwise complex expressions.
Derivative of Natural Log
Understanding the derivative of the natural log is a cornerstone in calculus because it frequently appears in growth models, physics equations, and more. The derivative of the natural log function, \( \ln(u) \), is \( \frac{1}{u} \) multiplied by the derivative of \( u \), represented as \( \frac{du}{dx} \).

This derivative formula is simple yet profound. It tells us how the natural log function changes with respect to changes in its input. For instance, in differentiating \( \ln(x^2 + 1) \), recognize the expression \( u = x^2 + 1 \). Thus, the derivative is \( \frac{1}{x^2 + 1} \cdot 2x \), thanks to the chain rule for differentiating \( u \).

This concept highlights the elegance of natural logarithms in calculus, making them indispensable for both practical and theoretical work. Grasping this foundation permits more complex differentiations involving logarithmic functions.

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Most popular questions from this chapter

Assume that \(N(t)\) denotes the size of a population at time \(t\) and that \(N(t)\) satisfies the differential equation $$ \frac{d N}{d t}=3 N\left(1-\frac{N}{20}\right) $$ Let \(f(N)=3 N\left(1-\frac{N}{20}\right)\) for \(N \geq 0\). Graph \(f(N)\) as a function of \(N\) and identify all equilibria (i.e., all points where \(\frac{d N}{d t}=0\) ).

Graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.) $$ y=|x-2| $$

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\left(1+\frac{1}{x}\right)^{1 / 4} \text { at } a=1 $$

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 \(x \pm|\Delta x|\). $$ f(x)=\sin x, x=-1 \pm 0.05 $$

Suppose that a certain plant is grown along a gradient ranging from nitrogen- poor to nitrogen-rich soil. Experimental data show that the average mass per plant grown in a soil with a total nitrogen content of \(1000 \mathrm{mg}\) nitrogen per \(\mathrm{kg}\) of soil is \(2.7 \mathrm{~g}\) and the rate of change of the average mass per plant at this nitrogen level is \(1.05 \times 10^{-3} \mathrm{~g}\) per mg change in total nitrogen per kg soil. Use a linear approximation to predict the average mass per plant grown in a soil with a total nitrogen content of \(1100 \mathrm{mg}\) nitrogen per \(\mathrm{kg}\) of soil.

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