Textbook Question
9–61. Evaluate and simplify y'.
y = (sin x / cos x+1)^1/3
3. Techniques of Differentiation
Derivatives of Trig Functions
5:27 minutesProblem 73a
Textbook Question
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 guidance1
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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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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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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