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Chapter 3: Problem 29

Find the equation of the tangent line to the curve \(y=(x-2)^{5}(x+1)^{2}\) at the point \((3,16)\).

Short Answer

Expert verified
y = 88x - 248

Step by step solution

01

- Find the derivative of the function

Given the function \(y=(x-2)^{5}(x+1)^{2}\). To find the equation of the tangent line, first find the derivative. Use the product rule: \(\frac{d}{dx}[u \times v] = u'v + uv'\), where \(u = (x-2)^5\) and \(v = (x+1)^2\).
02

- Differentiate each part

Differentiate \(u\) and \(v\) separately: \(u = (x-2)^5\) and \(v = (x+1)^2\). Thus: \(u' = 5(x-2)^{4}\) and \(v' = 2(x+1)\).
03

- Apply the product rule

Using the product rule, \(y' = u'v + uv'\). Substituting for \(u'\), \(v\), \(u\), and \(v'\): \(y' = 5(x-2)^{4}(x+1)^{2} + (x-2)^{5} \times 2(x+1)\).
04

- Simplify the derivative

Combine and simplify the terms: \(y' = 5(x-2)^{4}(x+1)^{2} + 2(x-2)^{5}(x+1)\).
05

- Evaluate the derivative at x=3

Substitute \(x=3\) into \(y'\): \(y'(3) = 5(3-2)^{4}(3+1)^{2} + 2(3-2)^{5}(3+1)\). This reduces to \(y'(3) = 5(1)^{4}(4)^{2} + 2(1)^{5}(4) = 80 + 8 = 88\).
06

- Use the point-slope form of the equation of a line

The point-slope form of a line is \(y - y_1 = m(x - x_1)\). Here, \((x_1, y_1) = (3, 16)\) and \(m = 88\). Substituting these into the equation: \(y - 16 = 88(x - 3)\).
07

- Simplify the equation

Expand and simplify the equation: \(y - 16 = 88x - 264\). Thus, \(y = 88x - 248\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
In calculus, the derivative represents how a function changes as its input changes. Think of it as the slope of the function at a given point. When you need to find the equation for a tangent line, the first step is to find the derivative of the function at the point where the tangent line touches the curve.
The function given is
\(y = (x-2)^{5}(x+1)^{2}\)
To find the derivative, we use rules like the product rule to differentiate the function correctly.
Product Rule
The product rule helps in finding the derivative of a product of two functions. The rule is stated as:
\[ \frac{d}{dx}[u \times v] = u'v + uv' \]
Here, \(u = (x-2)^5\) and \(v = (x+1)^2\).
We need to find the derivatives of \(u\) and \(v\) separately:
  • \(u = (x-2)^5\) so \(u' = 5(x-2)^4\)
  • \(v = (x+1)^2\) so \(v' = 2(x+1)\)

Now we can apply the product rule:
\[ y' = u'v + uv' \]
Substituting the values, we get:
\[ y' = 5(x-2)^{4}(x+1)^{2} + (x-2)^{5} \times 2(x+1) \]
Simplifying gives us the result:
\[ y' = 5(x-2)^{4}(x+1)^{2} + 2(x-2)^{5}(x+1) \]
This derivative tells us the slope of the function at any given \(x\).
Point-Slope Form
The point-slope form of the equation of a line is useful when you know the slope \(m\) of the line and a point \((x_1, y_1)\) on the line. The form is:
\[ y - y_1 = m(x - x_1) \]
In our case, the slope \(m\) is the value of the derivative at \(x = 3\), which we found to be 88. The point \((x_1, y_1)\) is \((3, 16)\).
Plugging these values into the point-slope form gives us:
\[ y - 16 = 88(x - 3) \]
Simplifying this equation, we get:
\[ y = 88x - 248 \]
This is the final tangent line equation. It shows the line that just touches the curve at the point \((3, 16)\).

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Most popular questions from this chapter

If \(f(x)\) is a function whose derivative is \(f^{\prime}(x)=\frac{1}{1+x^{2}}\), find the derivative of \(\frac{f(x)}{1+x^{2}}\).

Suppose that \(x\) and \(y\) are both differentiable functions of \(t\) and are related by the given equation. Use implicit differentiation with respect to \(t\) to determine \(\frac{d y}{d t}\) in terms of \(x, y\), and \(\frac{d x}{d t}\). $$x^{2} y^{2}=2 y^{3}+1$$

A function \(h(x)\) is defined in terms of a differentiable \(f(x)\). Find an expression for \(h^{\prime}(x)\). $$h(x)=2 f(2 x+1)$$

Suppose that \(x\) and \(y\) are related by the given equation and use implicit differentiation to determine \(\frac{d y}{d x}\). $$x^{2}-y^{2}=1$$

Suppose that \(x\) and \(y\) are related by the given equation and use implicit differentiation to determine \(\frac{d y}{d x}\). $$y^{5}-3 x^{2}=x$$
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