Logarithmic Differentiation: Derivative Of Complex Functions

In the realm of calculus, finding the derivative of a complex function can often feel like navigating a labyrinth. However, with the right tools and techniques, this challenge transforms into an elegant process. One such powerful technique is logarithmic differentiation. This method is particularly useful when dealing with functions that involve products, quotients, and powers, as it simplifies these operations by converting them into sums and differences. Today, we'll dive deep into how logarithmic differentiation works by applying it to a specific, rather intricate function: f(x)=rac{x^4(x-6)^7}{\left(x^2+6 ight)^4}. By the end of this discussion, you'll not only understand the mechanics of logarithmic differentiation but also appreciate its elegance in solving complex derivative problems, making those daunting mathematical expressions feel much more manageable and less intimidating.

Understanding Logarithmic Differentiation

Logarithmic differentiation is a technique used to find the derivative of a function by taking the natural logarithm of both sides of the equation before differentiating. This might seem like an extra step, but it's a strategic move that simplifies complex expressions. The core idea behind its effectiveness lies in the properties of logarithms. Recall that $\ln(ab) = \ln(a) + \ln(b)$, $\ln(a/b) = \ln(a) - \ln(b)$, and $\ln(a^n) = n\ln(a)$. When you apply the natural logarithm to a function involving multiplication, division, and exponentiation, these operations are transformed. Multiplication becomes addition, division becomes subtraction, and powers become coefficients. This transformation is crucial because differentiating sums and differences is generally much simpler than differentiating products and quotients. The process typically involves these steps: first, take the natural logarithm of both sides of the function; second, use logarithm properties to expand the expression; third, differentiate both sides with respect to $x$ (remembering to use the chain rule on the left side, resulting in $\frac{1}{y} \frac{dy}{dx}$ or $\frac{f'(x)}{f(x)}$); and finally, solve for $\frac{dy}{dx}$ (or $f'(x)$) by multiplying both sides by the original function $y$ (or $f(x)$). This systematic approach breaks down a complex problem into a series of simpler, manageable steps, making it a go-to method for many calculus students and professionals when faced with complicated functions.

Applying Logarithmic Differentiation to $f(x)=rac{x^4(x-6)7}{\left(x^2+6

ight)^4}$

Let's embark on the journey of finding the derivative of f(x)=rac{x^4(x-6)^7}{\left(x^2+6 ight)^4} using the powerful technique of logarithmic differentiation. Our first step is to set $y = f(x)$, so we have y=rac{x^4(x-6)^7}{\left(x^2+6 ight)^4}. Now, we take the natural logarithm of both sides: \ln(y) = \ln\left(\frac{x^4(x-6)^7}{\left(x^2+6 ight)^4}\right). The magic of logarithms comes into play here as we expand the right side using the properties $\ln(a/b) = \ln(a) - \ln(b)$ and $\ln(ab) = \ln(a) + \ln(b)$. This gives us: \ln(y) = \ln(x^4(x-6)^7) - \ln(\left(x^2+6 ight)^4). Further expanding the first term using $\ln(ab) = \ln(a) + \ln(b)$, we get: \ln(y) = \ln(x^4) + \ln((x-6)^7) - \ln(\left(x^2+6 ight)^4). Finally, we use the power rule for logarithms, $\ln(a^n) = n\ln(a)$, to bring down the exponents as coefficients: $\ln(y) = 4\ln(x) + 7\ln(x-6) - 4\ln(x^2+6)$. This is our simplified expression, ready for differentiation. The transformation from a complex quotient and product with powers into a sum and difference of simpler logarithmic terms is precisely why this method is so effective. It paves the way for straightforward differentiation in the next crucial step.

Differentiating Both Sides

Now that we have successfully transformed our complex function into a simpler logarithmic form, $\ln(y) = 4\ln(x) + 7\ln(x-6) - 4\ln(x^2+6)$, we can proceed to differentiate both sides with respect to $x$. This is where the chain rule becomes essential, especially on the left side. Differentiating $\ln(y)$ with respect to $x$ yields $\frac{1}{y} \frac{dy}{dx}$. For the right side, we differentiate each term individually. The derivative of $4\ln(x)$ is simply $4 \cdot \frac{1}{x} = \frac{4}{x}$. The derivative of $7\ln(x-6)$ requires the chain rule: $7 \cdot \frac{1}{x-6} \cdot \frac{d}{dx}(x-6) = \frac{7}{x-6} \cdot 1 = \frac{7}{x-6}$. Similarly, the derivative of $-4\ln(x^2+6)$ also requires the chain rule: $-4 \cdot \frac{1}{x^2+6} \cdot \frac{d}{dx}(x^2+6) = -4 \cdot \frac{1}{x^2+6} \cdot (2x) = \frac{-8x}{x^2+6}$. Putting it all together, the differentiated equation is: $\frac{1}{y} \frac{dy}{dx} = \frac{4}{x} + \frac{7}{x-6} - \frac{8x}{x^2+6}$. This step is critical as it links the derivative of the original function ($dy/dx$) to a combination of simpler rational functions. The use of the chain rule is paramount here, ensuring that we correctly account for the composite nature of the terms within the logarithms, particularly in the $(x-6)$ and $(x^2+6)$ components. Each derivative calculation is straightforward once the chain rule is applied correctly, reducing the complexity of the task significantly.

Solving for the Derivative $f'(x)$

Our equation at this stage is $\frac{1}{y} \frac{dy}{dx} = \frac{4}{x} + \frac{7}{x-6} - \frac{8x}{x^2+6}$. To find $f'(x)$, which is $\frac{dy}{dx}$, we need to isolate it. This is achieved by multiplying both sides of the equation by $y$. Remember that $y$ is our original function, f(x) = rac{x^4(x-6)^7}{\left(x^2+6 ight)^4}. So, we have: $\frac{dy}{dx} = y \left( \frac{4}{x} + \frac{7}{x-6} - \frac{8x}{x^2+6} \right)$. Substituting the original expression for $y$ back into the equation, we get: f'(x) = \frac{x^4(x-6)^7}{\left(x^2+6 ight)^4} \left( \frac{4}{x} + \frac{7}{x-6} - \frac{8x}{x^2+6} \right). While this is a correct expression for the derivative, it's often beneficial to simplify it further by finding a common denominator for the terms inside the parentheses and distributing the $y$ term. However, for many purposes, this form is perfectly acceptable and clearly shows the application of logarithmic differentiation. The final step of multiplying by the original function $y$ is what returns the derivative to its original form, undoing the logarithm and preparing it to be expressed in terms of $x$ only. This process elegantly resolves the complexities introduced by products, quotients, and powers in the initial function, making the derivative calculation tractable and accurate. The result is a precise representation of the rate of change of $f(x)$ with respect to $x$. This methodology ensures that even highly complex functions can be differentiated systematically.

Simplifying the Result (Optional but Recommended)

While the expression f'(x) = \frac{x^4(x-6)^7}{\left(x^2+6 ight)^4} \left( \frac{4}{x} + \frac{7}{x-6} - \frac{8x}{x^2+6} \right) is technically correct, it can be further simplified by combining the terms within the parentheses into a single fraction. This often leads to a more compact and sometimes more revealing form of the derivative. To do this, we find a common denominator for $\frac{4}{x}$, $\frac{7}{x-6}$, and $\frac{-8x}{x^2+6}$. The common denominator is $x(x-6)(x^2+6)$. Now, we rewrite each fraction with this common denominator:

• $\frac{4}{x} = \frac{4(x-6)(x^2+6)}{x(x-6)(x^2+6)}$
• $\frac{7}{x-6} = \frac{7x(x^2+6)}{x(x-6)(x^2+6)}$
• $\frac{-8x}{x^2+6} = \frac{-8x \cdot x(x-6)}{x(x-6)(x^2+6)} = \frac{-8x^2(x-6)}{x(x-6)(x^2+6)}$

Adding these together gives:

$\frac{4(x-6)(x^2+6) + 7x(x^2+6) - 8x^2(x-6)}{x(x-6)(x^2+6)}$

Let's expand the numerator:

$4(x^3 + 6x - 6x^2 - 36) + 7x^3 + 42x - 8x^3 + 48x^2$

$4x^3 - 24x^2 + 24x - 144 + 7x^3 + 42x - 8x^3 + 48x^2$

Combine like terms:

$(4x^3 + 7x^3 - 8x^3) + (-24x^2 + 48x^2) + (24x + 42x) - 144$

$3x^3 + 24x^2 + 66x - 144$

So, the expression inside the parentheses becomes:

$\frac{3x^3 + 24x^2 + 66x - 144}{x(x-6)(x^2+6)}$

Now, substitute this back into the expression for $f'(x)$:

f'(x) = \frac{x^4(x-6)^7}{\left(x^2+6 ight)^4} \cdot \frac{3x^3 + 24x^2 + 66x - 144}{x(x-6)(x^2+6)}

We can simplify this by canceling common factors. Notice that $x$ in the denominator cancels with one $x$ from $x^4$ in the numerator, leaving $x^3$. Also, $(x-6)$ in the denominator cancels with one $(x-6)$ from $(x-6)^7$ in the numerator, leaving $(x-6)^6$. Similarly, $(x^2+6)$ in the denominator cancels with one $(x^2+6)$ from $(x^2+6)^4$ in the numerator, leaving $(x^2+6)^3$. The simplified derivative is:

f'(x) = \frac{x^3(x-6)^6 (3x^3 + 24x^2 + 66x - 144)}{\left(x^2+6 ight)^3}

This simplified form, while requiring more algebraic manipulation, presents the derivative in a more consolidated manner. It's important to check for any potential common factors in the numerator's polynomial and the denominator's terms, though in this case, further simplification is not immediately apparent without factoring the cubic polynomial.

Conclusion: The Power of Logarithmic Differentiation

As we've navigated through the process of finding the derivative of f(x)=rac{x^4(x-6)^7}{\left(x^2+6 ight)^4}, the utility of logarithmic differentiation has become abundantly clear. This technique systematically breaks down complex functions involving products, quotients, and powers into simpler additive and subtractive logarithmic forms. By applying the properties of logarithms and then differentiating, we transformed a potentially arduous task into a series of manageable steps. The chain rule is an indispensable companion in this process, ensuring accuracy when differentiating composite logarithmic terms. Ultimately, we were able to find the derivative $f'(x)$ and even simplify it into a more compact form. This method not only provides an elegant solution but also underscores the interconnectedness of various calculus principles. Whether you are a student grappling with calculus homework or a professional seeking efficient methods for derivative calculations, mastering logarithmic differentiation will undoubtedly enhance your problem-solving toolkit. It's a testament to how understanding fundamental mathematical properties can unlock solutions to complex challenges.

For further exploration into differentiation techniques and calculus in general, you might find these resources helpful:

• Paul's Online Math Notes: A comprehensive resource for calculus, offering detailed explanations and examples on various topics, including differentiation. You can find it at https://tutorial.math.lamar.edu/.
• Khan Academy: Offers free online courses and practice exercises for calculus, providing a great platform for learning and reinforcing concepts. Check out their calculus section at https://www.khanacademy.org/math/calculus-1.