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Chapter 2: Problem 44

Find an equation for the tangent line to the graph at the specified value of \(x .\) $$ y=\sin \left(1+x^{3}\right), x=-3 $$

Short Answer

Expert verified
Slope is \(27\cos(-26)\), point is \((-3, \sin(-26))\), tangent line is \(y - \sin(-26) = 27\cos(-26) (x + 3)\)."

Step by step solution

01

Understand the Problem

We need to find the equation of the tangent line to the curve at a specific point. The general form of the equation for a tangent line is given by: \[y - y_1 = m(x - x_1)\] where \(m\) is the slope of the tangent line and \((x_1, y_1)\) is the point of tangency. Our function is \(y = \sin(1 + x^3)\) and we are evaluating it at \(x = -3\).
02

Differentiate the Function

To find the slope of the tangent line, we need to differentiate the function \(y = \sin(1 + x^3)\). Using the chain rule, \[\frac{dy}{dx} = \cos(1 + x^3) \cdot 3x^2\] This gives us the derivative of the function.
03

Evaluate the Derivative at the Given Point

Substitute \(x = -3\) into the derivative to find the slope of the tangent line at that point:\[m = \cos(1 + (-3)^3) \cdot 3(-3)^2 = \cos(-26) \cdot 27\] We calculate \(\cos(-26)\) and multiply by 27 to get the slope value.
04

Evaluate the Function at the Given Point

Substitute \(x = -3\) into the original function to find \(y_1\):\[y_1 = \sin(1 + (-3)^3) = \sin(-26)\] This gives us the y-coordinate of the point \((-3, y_1)\) on the curve.
05

Write the Equation of the Tangent Line

We use the point-slope form of the tangent line. With \(x_1 = -3\), \(y_1 = \sin(-26)\), and the slope \(m = \cos(-26) \cdot 27\):\[y - \sin(-26) = \cos(-26) \cdot 27 (x + 3)\] This equation gives the tangent line to the graph at \(x = -3\).

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Key Concepts

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

Chain Rule
The Chain Rule is a fundamental tool in calculus to differentiate composite functions. This rule is essential when you have a function nested inside another function.
To apply the Chain Rule, observe the pattern:
  • If you have a function of the form \( f(g(x)) \), differentiate the outer function \( f \) treating \( g(x) \) as a variable, then multiply by the derivative of the inner function \( g(x) \).
For this exercise, the function is \( y = \sin(1 + x^3) \), where:
  • The outer function is \( \sin(u) \), and the inner function is \( u = 1 + x^3 \).
The Chain Rule gives us the derivative of the function \( y \):
\[ \frac{dy}{dx} = \cos(1 + x^3) \cdot 3x^2 \]
The expression \( \cos(1 + x^3) \) is the derivative of \( \sin(u) \), and \( 3x^2 \) is the derivative of the inner function \( u = 1 + x^3 \).
Mastering the Chain Rule allows you to tackle complex differentiation problems with confidence.
Derivative
A derivative is a measure of how a function changes as its input changes. It represents the slope of the tangent line to the graph of the function.
Essentially, the derivative tells us the rate at which a function is changing at any given point.
  • In this particular problem, the goal is to find the slope of the tangent line at a specific point on the graph.
Once we've used the Chain Rule to find the derivative of \( y = \sin(1 + x^3) \):
\[ \frac{dy}{dx} = \cos(1 + x^3) \cdot 3x^2 \]
we substitute \( x = -3 \) into this derivative to calculate the slope \( m \).
The result tells us how steep the tangent line is at the point \((-3, y_1)\) on the curve. Knowing this rate of change is crucial for writing the tangent equation.
Point-Slope Form
The point-slope form is a useful formula for writing the equation of a line when you know the slope and one point on the line.
The formula is given by \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope and \((x_1, y_1)\) is the point on the line.
In this exercise:
  • The point of tangency is \((-3, \sin(-26))\).
  • The slope \( m \) is \( \cos(-26) \cdot 27 \).
Now you plug these values into the point-slope formula to get:\[ y - \sin(-26) = \cos(-26) \cdot 27 (x + 3) \]
Using this equation, you can describe precisely how the tangent line behaves relative to the curve at \( x = -3 \).
Point-slope form is an efficient and straightforward way to arrive at the equation of the tangent line.

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Most popular questions from this chapter

Find an equation for the tangent line to the graph at the specified value of \(x .\) $$ y=x \cos 3 x, x=\pi $$

Find a formula for $$ \frac{d}{d x}[f(g(h(x)))] $$

Find an equation for the tangent line to the graph at the specified value of \(x .\) $$ y=\sec ^{3}\left(\frac{\pi}{2}-x\right), x=-\frac{\pi}{2} $$

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0} .\) You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section \(4.8) .\) Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0},\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} f^{\prime}(x) .\) $$ \begin{array}{l}{\text { Let } \quad f(x)=\left\\{\begin{array}{ll}{x^{2}-16 x,} & {x<9} \\ {\sqrt{x},} & {x \geq 9}\end{array}\right.} \\ {\text { Is } f \text { continuous at } x=9 \text { ? Determine whether } f \text { is differ- }} \\ {\text { entiable at } x=9 . \text { If so, find the value of the derivative }} \\ {\text { there. }}\end{array} $$

Find \(f^{\prime}(x)\) $$ f(x)=\left(3 x^{2}+2 x-1\right)^{6} $$
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