Master Logarithmic Derivatives: Solved Examples

Hey guys, welcome back to the math corner! Today, we're diving deep into the awesome world of derivatives, specifically focusing on those tricky logarithmic functions. You know, the ones with the 'ln' in them? They can seem a bit intimidating at first, but trust me, once you get the hang of the rules, they're actually pretty cool to work with. We're going to tackle a bunch of examples, breaking them down step-by-step so you can confidently find the derivative of these functions. So, grab your notebooks, maybe a coffee, and let's get this done!

Understanding the Basics of Logarithmic Derivatives

Before we jump into the complex stuff, let's quickly recap why derivatives of logarithmic functions are important and the core rules we'll be using. The derivative of the natural logarithm, $f(x) = \ln(x)$, is simply $f'(x) = \frac{1}{x}$. This is our fundamental building block. When you have a more complex function inside the logarithm, like $f(x) = \ln(u(x))$, you need to apply the chain rule. The derivative then becomes $f'(x) = \frac{1}{u(x)} \cdot u'(x)$, or more commonly written as $f'(x) = \frac{u'(x)}{u(x)}$. This is super crucial, so keep it in mind! We'll also be using other standard derivative rules like the power rule ($(\frac{d}{dx} x^n = nx^{n-1}$)), the product rule ($(\frac{d}{dx} (uv) = u'v + uv')$), and the quotient rule ($(\frac{d}{dx} (\frac{u}{v}) = \frac{u'v - uv'}{v^2})$). Mastering these will make finding derivatives of logarithmic functions a breeze. Remember, practice is key, and by working through these examples, you'll build that muscle memory.

Example 1: Polynomials with Logarithms

Let's kick things off with a function that combines polynomial terms and logarithms: $f(x)=\ln ^2 x-2 \ln x+3$. This might look a little different because of the $\ln^2 x$ term. What that means is $(\ln x)^2$. So, to find the derivative, we'll treat $\ln x$ as our 'u' in the chain rule for that first term. The derivative of $(\ln x)^2$ will use the power rule first, bringing down the 2, and then we multiply by the derivative of $\ln x$, which is $\frac{1}{x}$. So, the derivative of $\ln^2 x$ is $2(\ln x)^1 \cdot \frac{1}{x} = \frac{2 \ln x}{x}$. For the second term, $-2 \ln x$, the derivative is straightforward: $-2 \cdot \frac{1}{x} = -\frac{2}{x}$. The constant term, 3, has a derivative of 0. Putting it all together, the derivative $f'(x)$ is $\frac{2 \ln x}{x} - \frac{2}{x} + 0$. We can simplify this by factoring out $\frac{2}{x}$ to get $f'(x) = \frac{2}{x}(\ln x - 1)$. See? Not too bad once you break it down. This example shows how the chain rule works with powers of logarithmic functions.

Example 2: Combinations of Polynomials and Logarithms

Alright, next up we have $f(x)=x^2+3 x-2-2 x \ln x$. This function has a mix of polynomial terms and a product involving $x$ and $\ln x$. We'll differentiate each part separately. The derivative of $x^2$ is $2x$. The derivative of $3x$ is $3$. The derivative of $-2$ is $0$. The tricky part here is the term $-2x \ln x$. This is a product of two functions: $-2x$ and $\ln x$. We need to use the product rule. Let $u = -2x$ and $v = \ln x$. Then $u' = -2$ and $v' = \frac{1}{x}$. Applying the product rule $(u'v + uv')$, we get $(-2)(\ln x) + (-2x)(\frac{1}{x})$. Simplifying this gives $-2 \ln x - 2$. Now, we combine the derivatives of all the terms: $f'(x) = 2x + 3 + 0 + (-2 \ln x - 2)$. Finally, let's simplify the expression by combining the constant terms: $f'(x) = 2x - 2 \ln x + 1$. This example is great for reinforcing the product rule when dealing with functions involving logarithms.

Example 3: The Quotient Rule with Logarithms

Now, let's tackle a function where we'll definitely need the quotient rule: $f(x)=\frac{x^2 \ln x}{x+1}$. Remember the quotient rule: $(\frac{d}{dx} (\frac{u}{v}) = \frac{u'v - uv'}{v^2})$. Here, our numerator is $u = x^2 \ln x$ and our denominator is $v = x+1$. First, let's find the derivative of the numerator, $u'$. This itself requires the product rule! For $x^2 \ln x$, let $a = x^2$ and $b = \ln x$. Then $a' = 2x$ and $b' = \frac{1}{x}$. So, $u' = a'b + ab' = (2x)(\ln x) + (x^2)(\frac{1}{x}) = 2x \ln x + x$. Now, let's find the derivative of the denominator, $v'$. This is simple: $v' = 1$. Now we can plug everything into the quotient rule formula:

$f'(x) = \frac{(2x \ln x + x)(x+1) - (x^2 \ln x)(1)}{(x+1)^2}$

Let's expand the numerator: $(2x^2 \ln x + 2x \ln x + x^2 + x) - x^2 \ln x$. Combining like terms in the numerator gives: $x^2 \ln x + 2x \ln x + x^2 + x$. So, the final derivative is:

$f'(x) = \frac{x^2 \ln x + 2x \ln x + x^2 + x}{(x+1)^2}$

We can factor out an $x$ from the first two terms in the numerator if we want to simplify further: $f'(x) = \frac{x(x \ln x + 2 \ln x) + x^2 + x}{(x+1)^2}$. Or even: $f'(x) = \frac{x((x+2)\ln x) + x(x+1)}{(x+1)^2}$. This example really tests your ability to combine multiple derivative rules.

Example 4: Logarithms of Fractions

Next up, we have $f(x)=\ln \left(\frac{x-1}{x+2}\right)$. Whenever you see a logarithm of a fraction, remember the logarithm property $\ln(\frac{a}{b}) = \ln a - \ln b$. This can simplify things immensely before you even start differentiating. So, we can rewrite our function as $f(x) = \ln(x-1) - \ln(x+2)$. Now, finding the derivative is much easier. The derivative of $\ln(x-1)$ is $\frac{1}{x-1} \cdot (1) = \frac{1}{x-1}$ (using the chain rule where $u = x-1$, so $u'=1$). The derivative of $\ln(x+2)$ is $\frac{1}{x+2} \cdot (1) = \frac{1}{x+2}$ (using the chain rule where $u = x+2$, so $u'=1$). So, $f'(x) = \frac{1}{x-1} - \frac{1}{x+2}$. To make this look nicer, we can combine the fractions by finding a common denominator:

$f'(x) = \frac{(x+2) - (x-1)}{(x-1)(x+2)}$

$f'(x) = \frac{x+2 - x+1}{x^2+x-2}$

$f'(x) = \frac{3}{x^2+x-2}$

Using logarithm properties before differentiating is a game-changer, guys! It often makes the process much more straightforward.

Example 5: Chain Rule with Powers and Logarithms

Now let's get a bit more complex with $f(x)=(2 x+\ln x)^3$. This is a classic chain rule scenario. Our 'outer' function is something cubed ($u^3$), and our 'inner' function is $u = 2x + \ln x$. First, apply the power rule to the outer function: the derivative of $u^3$ is $3u^2$. Then, we need to multiply by the derivative of the inner function, $u'$. The derivative of $2x + \ln x$ is $2 + \frac{1}{x}$. So, putting it all together, we get:

$f'(x) = 3(2 x+\ln x)^2 \cdot \left(2 + \frac{1}{x}\right)$

We can do a little simplification by distributing the $2 + \frac{1}{x}$ into the parenthesis, or combine it into a single fraction: $2 + \frac{1}{x} = \frac{2x+1}{x}$. So, the derivative can also be written as:

$f'(x) = 3(2 x+\ln x)^2 \frac{2x+1}{x}$

This problem is a fantastic illustration of how the chain rule is applied when you have a function raised to a power, and that function itself contains a logarithm.

Example 6: Quotient Rule with Logarithms (Again!)

Let's try another quotient rule problem, just to really drill it in: $f(x)=\frac{2 \ln x+x}{\ln x+1}$. Here, $u = 2 \ln x + x$ and $v = \ln x + 1$. Let's find the derivatives:

$u' = \frac{d}{dx}(2 \ln x + x) = 2 \cdot \frac{1}{x} + 1 = \frac{2}{x} + 1 = \frac{2+x}{x}$

$v' = \frac{d}{dx}(\ln x + 1) = \frac{1}{x} + 0 = \frac{1}{x}$

Now, apply the quotient rule $f'(x) = \frac{u'v - uv'}{v^2}$:

$f'(x) = \frac{(\frac{2+x}{x})(\ln x+1) - (2 \ln x+x)(\frac{1}{x})}{(\ln x+1)^2}$

Let's simplify the numerator. Multiply the first term:

$(\frac{2+x}{x})(\ln x+1) = \frac{(2+x)\ln x + (2+x)}{x} = \frac{2 \ln x + x \ln x + 2 + x}{x}$

Now, subtract the second part of the numerator:

$\frac{2 \ln x + x \ln x + 2 + x}{x} - \frac{2 \ln x+x}{x}$

This gives:

$\frac{2 \ln x + x \ln x + 2 + x - (2 \ln x+x)}{x} = \frac{2 \ln x + x \ln x + 2 + x - 2 \ln x - x}{x}$

$= \frac{x \ln x + 2}{x}$

So, the derivative is:

$f'(x) = \frac{\frac{x \ln x + 2}{x}}{(\ln x+1)^2} = \frac{x \ln x + 2}{x(\ln x+1)^2}$

Whew! This one involved a lot of algebraic manipulation, but breaking it down piece by piece makes it manageable.

Example 7: Nested Logarithms

Finally, let's look at $f(x)=\ln (x+\ln x)$. This is another direct application of the chain rule. Our 'outer' function is $\ln(u)$, and our 'inner' function is $u = x + \ln x$. The derivative of $\ln(u)$ is $\frac{1}{u}$. The derivative of our inner function $u = x + \ln x$ is $u' = 1 + \frac{1}{x}$.

So, applying the chain rule, $f'(x) = \frac{1}{u} \cdot u'$:

$f'(x) = \frac{1}{x+\ln x} \cdot \left(1 + \frac{1}{x}\right)$

We can combine the terms in the parenthesis into a single fraction: $1 + \frac{1}{x} = \frac{x+1}{x}$. So the final derivative is:

$f'(x) = \frac{1}{x+\ln x} \cdot \frac{x+1}{x}$

$f'(x) = \frac{x+1}{x(x+\ln x)}$

This type of function, with nested logarithms, is common and understanding the chain rule is paramount here. You just peel back the layers one by one!

Wrapping Up

And there you have it, folks! We've gone through several examples of finding derivatives of logarithmic functions, covering everything from basic combinations to more complex scenarios involving the product rule, quotient rule, and chain rule. The key takeaways are: always remember the basic derivative of $\ln x$ is $\frac{1}{x}$, apply the chain rule diligently when you have functions within functions, and don't shy away from simplifying using logarithm properties or algebraic manipulation. Practice these types of problems, and soon you'll be a derivative pro! Let me know in the comments if you have any questions or want to see more examples. Keep practicing, and I'll see you in the next one!