Interactive Derivative Calculation

Introduction

Welcome! This guide breaks down the differentiation of the function $ f(x) = \frac{xe^{x}}{\ln(x^{2})} $. Each step from the provided solution is explained, and you can click to view the definitions of the calculus rules used. Follow along to understand how the derivative is found.

Step 1: Identify the main rule to apply.

The function $f(x)$ is a fraction. This means we need to start with the Quotient Rule to find its derivative.

The Quotient Rule states that if $ f(x) = \frac{u(x)}{v(x)} $, then its derivative $ f'(x) $ is:

$$ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

For our function $ f(x) = \frac{xe^{x}}{\ln(x^{2})} $, we define:

Numerator: $ u(x) = xe^{x} $
Denominator: $ v(x) = \ln(x^{2}) $

The next steps will be to find the derivatives of $u(x)$ and $v(x)$, which are $u'(x)$ and $v'(x)$ respectively.

Step 2: Find the derivative of the numerator, $u'(x)$.

The numerator is $ u(x) = xe^{x} $. This is a product of two functions, $g(x) = x$ and $h(x) = e^x$. Thus, we use the Product Rule.

The Product Rule states that if $ u(x) = g(x)h(x) $, then its derivative $ u'(x) $ is:

$$ u'(x) = g'(x)h(x) + g(x)h'(x) $$

Here, let $g(x) = x$ and $h(x) = e^{x}$. Their derivatives are:

$ g'(x) = \frac{d}{dx}(x) = 1 $
$ h'(x) = \frac{d}{dx}(e^{x}) = e^{x} $

Applying the product rule for $u'(x)$:

$$ u'(x) = (1)(e^{x}) + (x)(e^{x}) $$ $$ u'(x) = e^{x} + xe^{x} $$ $$ u'(x) = e^{x}(1+x) $$

So, the derivative of the numerator is $ u'(x) = e^{x}(1+x) $.

Step 3: Find the derivative of the denominator, $v'(x)$.

The denominator is $ v(x) = \ln(x^{2}) $. We can find its derivative using a property of logarithms or by applying the Chain Rule.

Method 1: Using Logarithm Properties

First, simplify $ \ln(x^2) $ using the property $ \ln(a^b) = b\ln(a) $:

$$ v(x) = \ln(x^{2}) = 2\ln(x) $$

Now, find the derivative of $ v(x) = 2\ln(x) $:

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

Method 2: Using the Chain Rule Directly (Alternative)

Alternatively, for $ v(x) = \ln(x^{2}) $, let the outer function be $ k(w) = \ln(w) $ and the inner function be $ w(x) = x^2 $. The Chain Rule is then applied.

The Chain Rule states that if $ v(x) = k(w(x)) $, then its derivative $ v'(x) $ is:

$$ v'(x) = k'(w(x)) \cdot w'(x) $$ Or, in Leibniz notation: $ \frac{dv}{dx} = \frac{dk}{dw} \cdot \frac{dw}{dx} $

The derivatives are:

$ \frac{dk}{dw} = \frac{d}{dw}(\ln(w)) = \frac{1}{w} $
$ \frac{dw}{dx} = \frac{d}{dx}(x^2) = 2x $

Applying the Chain Rule:

$$ v'(x) = \frac{1}{w(x)} \cdot 2x = \frac{1}{x^2} \cdot 2x = \frac{2x}{x^2} = \frac{2}{x} $$

Both methods correctly give $ v'(x) = \frac{2}{x} $.

Step 4: Apply the Quotient Rule formula.

Now we have all the necessary components to use the Quotient Rule. Let's list them:

$ u(x) = xe^{x} $
$ v(x) = \ln(x^{2}) $
$ u'(x) = e^{x}(1+x) $
$ v'(x) = \frac{2}{x} $

Recall the Quotient Rule formula: $ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $

Substitute these into the formula:

$$ f'(x) = \frac{[e^{x}(1+x)] \cdot [\ln(x^{2})] - [xe^{x}] \cdot \left[\frac{2}{x}\right]}{[\ln(x^{2})]^2} $$

This expression is the derivative, but it can be simplified further.

Step 5: Simplify the expression.

Let's simplify the numerator of the expression obtained in Step 4.

First, simplify the term $ [xe^{x}] \cdot \left[\frac{2}{x}\right] $:

$$ [xe^{x}] \cdot \left[\frac{2}{x}\right] = 2e^{x} $$ (The $x$ in the numerator and denominator cancel out.)

So, the numerator $ [e^{x}(1+x)] \cdot [\ln(x^{2})] - [xe^{x}] \cdot \left[\frac{2}{x}\right] $ becomes:

$$ e^{x}(1+x)\ln(x^{2}) - 2e^{x} $$

Next, factor out the common term $e^{x}$ from the numerator:

$$ e^{x}[(1+x)\ln(x^{2}) - 2] $$

Now consider the denominator: $ [\ln(x^{2})]^2 $. Using the property $ \ln(x^2) = 2\ln(x) $, the denominator can also be written as:

$$ (2\ln(x))^2 = 4(\ln(x))^2 $$

Combining the simplified numerator and the original denominator, the derivative $f'(x)$ is:

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

Or, using the alternative form of the denominator:

$$ f'(x) = \frac{e^{x}[(1+x)\ln(x^{2}) - 2]}{4(\ln(x))^2} $$

Both forms are equivalent and represent the simplified derivative.

Final Result

This is the final derivative of the given function, as derived in the provided solution: