Problem 64 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to \(y=\) \(\left(x^{2}+1\right)^{3}\left(x^{4}+1\right)^{2}\) at \((1,32)\).

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = 176x - 144\).

Step 1: Understand the Function

Given the function \(y = (x^2 + 1)^3 (x^4 + 1)^2\), we are tasked with finding the tangent line at the point \((1, 32)\). To do this, we need to find the derivative of the function to determine the slope of the tangent line.

Step 2: Use the Product Rule

The function is a product of \((x^2 + 1)^3\) and \((x^4 + 1)^2\). Use the product rule: \((uv)' = u'v + uv'\). Let \(u = (x^2 + 1)^3\) and \(v = (x^4 + 1)^2\). This means we need to find \(u'\) and \(v'\).

Step 3: Differentiate Using the Chain Rule

First, find \(u'\). For \(u = (x^2 + 1)^3\):- Use the chain rule: Derivative of the outer function: \(3(x^2+1)^2\).- Derivative of the inside function \(x^2+1\) is \(2x\).- So, \(u' = 3(x^2+1)^2 \cdot 2x = 6x(x^2+1)^2\).Next, find \(v'\). For \(v = (x^4 + 1)^2\):- Use the chain rule: Derivative of the outer function: \(2(x^4+1)\).- Derivative of the inside function \(x^4+1\) is \(4x^3\).- So, \(v' = 2(x^4+1) \cdot 4x^3 = 8x^3(x^4+1)\).

Step 4: Apply the Product Rule

Substitute \(u\), \(v\), \(u'\), and \(v'\) into the product rule:\[y' = (6x(x^2+1)^2)(x^4+1)^2 + (x^2+1)^3(8x^3(x^4+1))\]

Step 5: Evaluate the Derivative at the Point

We need the slope at \((1, 32)\). Substitute \(x = 1\) into the expression obtained from the product rule:- \(u(1) = (1^2 + 1)^3 = 8\)- \(v(1) = (1^4 + 1)^2 = 4\)- \(u'(1) = 6 \cdot 1 \cdot (1^2 + 1)^2 = 12\)- \(v'(1) = 8 \cdot 1^3 \cdot (1^4 + 1) = 16\)Substitute into the formula:\[y'(1) = 12 \cdot 4 + 8 \cdot 16 = 48 + 128 = 176\]

Step 6: Find the Equation of the Tangent Line

The tangent line at a point \((x_0, y_0)\) with slope \(m\) is given by \(y - y_0 = m(x - x_0)\). We have \((x_0, y_0) = (1, 32)\) and \(m = 176\). Substitute these values:\[y - 32 = 176(x - 1)\]Simplify:\[y = 176x - 144\]

Step 7: Final Answer

The equation of the tangent line to the curve at the point \((1, 32)\) is \(y = 176x - 144\).

Key Concepts

Product Rule

When dealing with functions that are products of two simpler functions, the product rule is a vital tool in calculus. It helps us differentiate these complex functions effectively. In mathematical form, the product rule states:

• If you have a function that is the product of two functions, say \( y = uv \), then its derivative \( y' \) is given by \( y' = u'v + uv' \).
• Here, \( u' \) and \( v' \) are the derivatives of the individual functions \( u \) and \( v \).

By applying the product rule, we're able to systematically find the derivative of a product of functions and thus solve more intricate problems such as finding tangent lines at specific points. In our exercise, the functions \( u = (x^2 + 1)^3 \) and \( v = (x^4 + 1)^2 \) require this very strategy.

Chain Rule

The chain rule is another essential concept that comes into play, especially when dealing with composite functions. A composite function is basically a function nested inside another function.

• To differentiate a composite function, we use the chain rule, defined as \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).
• Think of it as differentiating the outer function first, and then multiplying by the derivative of the inner function.

For instance, in our problem:

• The function \( u = (x^2 + 1)^3 \) is differentiated using the chain rule to find \( u' = 3(x^2+1)^2 \cdot 2x \).
• Similarly, \( v = (x^4 + 1)^2 \) yields \( v' = 2(x^4+1) \cdot 4x^3 \).

Using the chain rule simplifies the process of finding derivatives of more elaborate functions, which is crucial for accurate calculus problem-solving.

Derivatives

Derivatives play a pivotal role in many calculus problems, particularly in finding slopes and rates of change. They tell you how a function's value changes as its input changes.

• The derivative at a certain point gives you the slope of the tangent line to the curve at that point.
• This helps in understanding how steep or flat the curve is at that particular location.

In the exercise provided, we calculated the derivative of the function \( y = (x^2 + 1)^3 (x^4 + 1)^2 \) using both the product rule and chain rule, leading us to find the slope at the specific point \((1, 32)\). Understanding and computing derivatives is crucial to tackling problems that involve tangent lines and changes in functions.

Calculus Problem Solving

Effective calculus problem solving often combines several techniques and rules. To tackle a calculus problem, such as finding the equation of a tangent line, one typically follows a structured approach:

• First, understand the given function and what is being asked. In our case, the task was to find the tangent line at a certain point.
• Identify if there are any products or composite functions involved and choose the appropriate rules, like the product and chain rules, for differentiation.
• Differentiate the function correctly to find the derivative, which gives you the slope needed for the tangent line.
• Substitute the specific point into the derivative to find the slope there.
• Finally, use the point-slope form of the tangent line equation to find the final solution.

By systematically combining these steps, one can solve calculus problems that appear complex at first glance, achieving clear and precise solutions.