Question

Use the properties of logarithms to simplify the following functions before computing $f^{\prime }(x)$. $$f(x)=\ln \frac{2x}{{(x^2+1)}^3}$$

Short answer

Question: Find the simplified derivative of the function $f(x)=\ln \frac{2x}{{(x^2+1)}^3}$. Answer: The simplified derivative of the given function is $f^{\prime }(x)=\frac{1}{x}-\frac{6x}{x^2+1}$.

Step-by-step solution

Use logarithmic properties to simplify the function

In order to simplify the given function, we can use the following log properties: $$\ln (ab)=\ln (a)+\ln (b)$$ $$\ln \left(\frac{a}{b}\right)=\ln (a)-\ln (b)$$ $$\ln (a^n)=n\ln (a)$$ Using these properties, we can rewrite the given function as: $$f(x)=\ln \frac{2x}{{(x^2+1)}^3}=\ln (2x)-\ln \left({(x^2+1)}^3\right)$$ Now, apply the third property, $$f(x)=\ln (2x)-3\ln (x^2+1)$$

Compute the derivative of the simplified function

Now that we have our simplified function, let's compute its derivative with respect to x. Recall that the derivative of the natural logarithm is given by: $$\frac{d}{dx}\ln (a)=\frac{\frac{d}{dx}(a)}{a}$$ Using the chain rule, we can compute the derivative of $f(x)$ as follows: $$f^{\prime }(x)=\frac{d}{dx}[\ln (2x)-3\ln (x^2+1)]$$ Let's compute these derivatives separately: For $\ln (2x)$, we have: $$\frac{d}{dx}\ln (2x)=\frac{\frac{d}{dx}(2x)}{2x}=\frac{2}{2x}=\frac{1}{x}$$ For $\ln (x^2+1)$, we have: $$\frac{d}{dx}\ln (x^2+1)=\frac{\frac{d}{dx}(x^2+1)}{x^2+1}=\frac{2x}{x^2+1}$$ Now, substitute these derivatives back into the original derivative expression: $$f^{\prime }(x)=\frac{1}{x}-3\left(\frac{2x}{x^2+1}\right)=\frac{1}{x}-\frac{6x}{x^2+1}$$ So, the simplified derivative of the given function is: $$f^{\prime }(x)=\frac{1}{x}-\frac{6x}{x^2+1}$$

Key concepts

Chain rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. Simply put, when you have a function nested inside another function, the chain rule helps in finding the derivative of the entire expression.
For instance, consider a function such as $h(x)=(3x+2{)}^5$. Here, the inside function is $u(x)=3x+2$, and the outside function is $v(u)=u^5$. The chain rule states that the derivative of $h(x)$ is given by the derivative of $v$ with respect to $u$ times the derivative of $u$ with respect to $x$.
In mathematical terms:

• $h^{\prime }(x)=v^{\prime }(u)\cdot u^{\prime }(x)$
• $v^{\prime }(u)=5u^4$
• $u^{\prime }(x)=3$
• Thus, $h^{\prime }(x)=5(3x+2{)}^4\cdot 3$
• Simplifying gives: $15(3x+2{)}^4$

This process applies to any composite functions like $f(x)=\ln (g(x))$, where the chain rule aids in deriving $f^{\prime }(x)=\frac{g^{\prime }(x)}{g(x)}$. This is crucial when working with complex logarithmic functions, enabling the derivation of their derivatives step by step.

Derivative simplification

Simplifying derivatives is an essential part of solving calculus problems accurately and efficiently. Once you've calculated a derivative, it might look complex or cumbersome. Simplifying it involves reducing the expression to its simplest form.
This is done by performing algebraic manipulations such as combining like terms, canceling out factors, and performing fractional reductions wherever applicable.
For example, let's say you derive an expression like $f^{\prime }(x)=\frac{2x}{x^2+1}-\frac{3}{x}$. Simplifying involves identifying common denominators or terms. In this case, you can re-write the first term with a common denominator:
$f^{\prime }(x)=\frac{2x^2-3(x^2+1)}{x(x^2+1)}$. Upon simplification, the terms within the numerator can be combined and reduced efficiently.

• With basic algebra, the numerator $2x^2-3x^2-3$ simplifies to $-x^2-3$.
• The whole derivative now reads $\frac{-x^2-3}{x(x^2+1)}$.

It's important to note that simplifying makes the derivative easier to interpret and functionally opportune for further calculations such as integrations or evaluations at specific points.

Natural logarithms

Natural logarithms, denoted $\ln (x)$, are logarithms to the base $e$, where $e\approx 2.7182818$ is a mathematical constant. They are fundamental in both pure and applied mathematics, due to their unique properties.
Natural logarithms transform multiplicative processes into additive ones, thus simplifying the complexity of differentiation and integration of log functions.
Here are key properties that are widely utilized:

• The derivative of $\ln (x)$ is $\frac{1}{x}$. This means if $f(x)=\ln (x)$, then $f^{\prime }(x)=\frac{1}{x}$.
• The chain rule can be applied whenever $\ln$ is part of a composition, like $\ln (u(x))$. The derivative is $\frac{u^{\prime }(x)}{u(x)}$.

Natural logarithms also simplify computations associated with exponential growth or decay processes. They are ubiquitous in calculus, especially involving the derivatives of logarithmic functions. This makes natural logarithms a vital tool for solving and simplifying calculus problems, particularly when working with exponential models or growth decay applications.