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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 73a
Chapter 3, Problem 73a

Find an equation of the line tangent to the following curves at the given value of x.
y = 1+2 sin x; x = π/6

Verified step by step guidance
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First, identify the function for which we need to find the tangent line. The function given is \( y = 1 + 2 \sin x \).
To find the equation of the tangent line, we need the slope of the tangent line at \( x = \frac{\pi}{6} \). This requires finding the derivative of the function, \( y = 1 + 2 \sin x \).
Differentiate the function with respect to \( x \). The derivative of \( y = 1 + 2 \sin x \) is \( y' = 2 \cos x \).
Evaluate the derivative at \( x = \frac{\pi}{6} \) to find the slope of the tangent line. Substitute \( x = \frac{\pi}{6} \) into \( y' = 2 \cos x \) to get the slope.
Finally, use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope found in the previous step and \( (x_1, y_1) \) is the point on the curve at \( x = \frac{\pi}{6} \). Calculate \( y_1 \) by substituting \( x = \frac{\pi}{6} \) into the original function \( y = 1 + 2 \sin x \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line represents the instantaneous rate of change of the function at that point, which can be found using the derivative of the function.
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Slopes of Tangent Lines

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. For the function y = 1 + 2 sin x, the derivative will provide the slope of the tangent line at any point x.
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Derivatives

Evaluating Functions

Evaluating a function involves substituting a specific value into the function to find the corresponding output. In this case, to find the equation of the tangent line at x = π/6, we need to evaluate both the function and its derivative at this point to determine the y-coordinate and the slope of the tangent line.
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Evaluating Composed Functions
Related Practice
Textbook Question

Tangent lines Suppose f(2)=2 and f′(2) =3. Let g(x) = x²f(x) and h(x) = f(x) / x−3.

b. Find an equation of the line tangent to y = h(x) at x=2.

Textbook Question

The right-sided and left-sided derivatives of a function at a point aa are given by f+′(a)=limh→0+f(a+h)−f(a)hf_{+}^{\(\prime\)}\(\left\)(a\(\right\))={\(\displaystyle\]\lim\)_{h\(\to\)0^{+}}{\(\frac{f(a+h)-f(a)}{h}\)}} and f−′(a)=limh→0−f(a+h)−f(a)hf_{-}^{\(\prime\)}\(\left\)(a\(\right\))={\(\displaystyle\]\lim\)_{h\(\to\)0^{-}}{\(\frac{f(a+h)-f(a)}{h}\)}}, respectively, provided these limits exist. The derivative f′(a)f^{\(\prime\)}\(\left\)(a\(\right\)) exists if and only if f+′(a)=f−′(a)f_{+}^{\(\prime\)}\(\left\)(a\(\right\))=f_{-}^{\(\prime\)}\(\left\)(a\(\right\)).

Compute f+′(a)f_{+}^{\(\prime\)}\(\left\)(a\(\right\)) and f−′(a)f_{-}^{\(\prime\)}\(\left\)(a\(\right\)) at the given point aa.

f(x)={4−x2 if x≤12x+1 if x>1f(x)=\(\begin{cases}\)4-x^2~\(\text{if}\)~x\(\leq{1}\)\\2x+1~\(\text{if}\)~x\(\gt{1}\]\end{cases}\); a=1a=1

Textbook Question

Calculate the derivative of the following functions.

y = (p+3)² sin p²

Textbook Question

27–76. Calculate the derivative of the following functions.

y = √f(x), where f is differentiable and nonnegative at x.

Textbook Question

First and second derivatives Find f′(x),f′′(x).

f(x) = x/x+2

Textbook Question

Suppose f(3) = 1 and f′(3) = 4. Let g(x) = x2 + f(x) and h(x) = 3f(x).

Find an equation of the line tangent to y = g(x) at x = 3.

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