Use formulas (1) and (2) and the power rule to find the derivatives of the following functions. f ( x ) = x 2 7 . Derivative of Linear Function If f ( x ) = m x + b , then we have f ' ( x ) = m . ( 1 ) Constant Rule The derivative of a constant function f ( x ) = b is zero. That is, f ' ( x ) = 0 . ( 2 ) Power Rule Let r be any number, and let f ( x ) = x r . Then f ' ( x ) = r x r − 1 .
Use formulas (1) and (2) and the power rule to find the derivatives of the following functions. f ( x ) = x 2 7 . Derivative of Linear Function If f ( x ) = m x + b , then we have f ' ( x ) = m . ( 1 ) Constant Rule The derivative of a constant function f ( x ) = b is zero. That is, f ' ( x ) = 0 . ( 2 ) Power Rule Let r be any number, and let f ( x ) = x r . Then f ' ( x ) = r x r − 1 .
Solution Summary: The author explains how to determine the derivative of the given function using the power rule.
Points z1 and z2 are shown on the graph.z1 is at (4 real,6 imaginary), z2 is at (-5 real, 2 imaginary)Part A: Identify the points in standard form and find the distance between them.Part B: Give the complex conjugate of z2 and explain how to find it geometrically.Part C: Find z2 − z1 geometrically and explain your steps.
A polar curve is represented by the equation r1 = 7 + 4cos θ.Part A: What type of limaçon is this curve? Justify your answer using the constants in the equation.Part B: Is the curve symmetrical to the polar axis or the line θ = pi/2 Justify your answer algebraically.Part C: What are the two main differences between the graphs of r1 = 7 + 4cos θ and r2 = 4 + 4cos θ?
A curve, described by x2 + y2 + 8x = 0, has a point A at (−4, 4) on the curve.Part A: What are the polar coordinates of A? Give an exact answer.Part B: What is the polar form of the equation? What type of polar curve is this?Part C: What is the directed distance when Ø = 5pi/6 Give an exact answer.
Chapter 1 Solutions
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Differential Equation | MIT 18.01SC Single Variable Calculus, Fall 2010; Author: MIT OpenCourseWare;https://www.youtube.com/watch?v=HaOHUfymsuk;License: Standard YouTube License, CC-BY