Question

Use logarithmic differentiation to find \(\frac{d y}{d x}\), then find the equation of the tangent line at the indicated \(x\) -value. $$y=\frac{(x+1)(x+2)}{(x+3)(x+4)}, \quad x=0$$

Short answer

The equation of the tangent line at \( x = 0 \) is \( y = \frac{7}{72}x + \frac{1}{6} \).

Step-by-step solution

Express the Function as a Logarithm

We begin with the function \( y = \frac{(x+1)(x+2)}{(x+3)(x+4)} \). To differentiate using logarithmic differentiation, take the natural logarithm of both sides: \( \ln y = \ln \left( \frac{(x+1)(x+2)}{(x+3)(x+4)} \right) \). Using properties of logarithms, this can be expanded to \( \ln y = \ln(x+1) + \ln(x+2) - \ln(x+3) - \ln(x+4) \).

Differentiate Both Sides

Differentiate both sides with respect to \(x\). The derivative of \( \ln y \) is \( \frac{1}{y} \frac{d y}{d x} \) (by the chain rule), so \( \frac{1}{y} \frac{d y}{d x} = \frac{1}{x+1} + \frac{1}{x+2} - \frac{1}{x+3} - \frac{1}{x+4} \).

Solve for \( \frac{d y}{d x} \)

Multiply through by \(y\) to isolate \( \frac{d y}{d x} \): \( \frac{d y}{d x} = y \left( \frac{1}{x+1} + \frac{1}{x+2} - \frac{1}{x+3} - \frac{1}{x+4} \right) \). Substitute \( y = \frac{(x+1)(x+2)}{(x+3)(x+4)} \) back in: \( \frac{d y}{d x} = \frac{(x+1)(x+2)}{(x+3)(x+4)} \left( \frac{1}{x+1} + \frac{1}{x+2} - \frac{1}{x+3} - \frac{1}{x+4} \right) \).

Evaluate \( \frac{d y}{d x} \) at \( x = 0 \)

Substitute \( x = 0 \) into the expression for \( \frac{d y}{d x} \). The expression becomes \( \frac{(0+1)(0+2)}{(0+3)(0+4)} \left( \frac{1}{0+1} + \frac{1}{0+2} - \frac{1}{0+3} - \frac{1}{0+4} \right) \). Simplifying, we get \( \frac{2}{12} \left( 1 + \frac{1}{2} - \frac{1}{3} - \frac{1}{4} \right) = \frac{1}{6} \times \frac{7}{12} = \frac{7}{72} \).

Find the Equation of the Tangent Line

The equation of the tangent line at \( x = 0 \) is of the form \( y - y_0 = m(x - x_0) \), where \( m \) is the derivative \( \frac{d y}{d x} \) at \( x = 0 \). First, calculate \( y_0 \) by evaluating \( y \) at \( x = 0 \): \( y_0 = \frac{(0+1)(0+2)}{(0+3)(0+4)} = \frac{2}{12} = \frac{1}{6} \). The slope \( m = \frac{7}{72} \). Therefore, the equation is \( y - \frac{1}{6} = \frac{7}{72}(x - 0) \) or \( y = \frac{7}{72}x + \frac{1}{6} \).

Key concepts

Tangent Line Equation

Finding the tangent line equation is a task where we determine the line that just "touches" a curve at a specific point without crossing it. For a curve described by a function, the tangent line at any given point indicates the function's instantaneous rate of change at that point. To find the tangent line:

• Identify the point at which you want the tangent line. Here, it's where the given function is evaluated at a point, such as when a specific value of \(x\) is substituted into the function.
• Compute the function's value (\(y_0\)) and its slope (\(m\)) at this point. In our example, the slope \(m\) is found to be \(\frac{7}{72}\).
• Once the point and slope are known, use the point-slope form: \(y - y_0 = m(x - x_0)\) to construct the equation of the tangent line.

Once these values are substituted into the point-slope formula, the equation is easily written down, showing students how to transition from differentiation tasks to geometric interpretation. This makes the abstract concept more "real-world".

Chain Rule

The chain rule is an essential differentiation technique in calculus, allowing us to find the derivative of a composite function. This is crucial when functions are nested within each other. The chain rule states that if a function \(y\) is the composition of an inner function \(u=g(x)\) and an outer function \(f\), so that \(y=f(u)\), then the derivative \(\frac{dy}{dx}\) is given by multiplying the derivative of the outer function by the derivative of the inner function. In mathematical terms:\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]When using logarithmic differentiation, you often implicitly use the chain rule, as seen when differentiating \(\ln y\). Here, we used the rule to differentiate \(\ln y\) with respect to \(x\), resulting in \(\frac{1}{y} \frac{dy}{dx}\). Employing the chain rule correctly is like being handed the keys to untangling even the most complex functions into manageable pieces.

Properties of Logarithms

The properties of logarithms are fundamental tools that simplify complex expressions, especially when differentiating functions. Logarithms transform multiplicative relationships into additive ones and division into subtraction, which can make differentiating products and quotients significantly easier. Some key properties that are incredibly helpful include:

• \(\ln(ab) = \ln a + \ln b\): Turns multiplication into addition.
• \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\): Converts division into subtraction.
• \(\ln a^n = n \ln a\): Allows moving exponents to the front as a multiplier.

In the original exercise, these properties were crucial. They helped transform the complex original function \(y = \frac{(x+1)(x+2)}{(x+3)(x+4)}\) into a simpler form: a series of logarithmic terms. This simplification allows for more straightforward differentiation.

Implicit Differentiation

Implicit differentiation is used when a function is not expressed in the standard "\(y=f(x)\)" form. It allows us to differentiate functions where \(y\) is intermingled with \(x\). In our scenario, implicit differentiation manages to handle the multi-layered approach needed when parts of the function are compounded, or where \(y\) is mixed with other forms. The process involves:

• Differentiating each side of the equation with respect to \(x\).
• Applying the chain rule wherever \(y\) or implicit compositions appear.
• Solving for \(\frac{dy}{dx}\), which might involve algebraic manipulation if \(y\) terms are left on both sides.

In our exercise, applying implicit differentiation following logarithmic transformation ensures that we can efficiently find \(\frac{d y}{d x}\) even when the path isn't straightforward. This technique is especially handy in contexts where explicit solutions are complex or impossible to isolate directly.