2.4 derivatives of trigonometric functions. example 1 differentiate y = x 2 sin x. solution: using...
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Example 2
An object at the end of a vertical spring is stretched 4 cm beyond its rest position and released at time t = 0 (note that the downward direction is positive.)
Its position at time t is
s = f (t) = 4 cos t
Find the velocity and acceleration
at time t and use them to analyze
the motion of the object.
Example 2 – Solution
The object oscillates from the lowest point (s = 4 cm) to the highest point (s = –4 cm). The period of the oscillation is 2, which is the period of cos t.
cont’d
Example 3 – Solution
• The speed is | v | = 4 | sin t |, which is greatest when | sin t | = 1, that is, when cos t = 0.
• So the object moves fastest as it passes through its equilibrium position (s = 0). Its speed is 0 when sin t = 0, that is, at the high and low points.
• The acceleration a = –4 cos t = 0 when s = 0. It has greatest magnitude at the high and low points.
cont’d
Example
• Find F '(x) if F (x) = .
Solution: (using the first definition)• F (x) = (f g)(x) = f (g(x)) where f (u) = and g (x) = x2 + 1.
• Since and g(x) = 2x
we have F (x) = f (g (x)) g (x)
Example• Differentiate y = (x3 – 1)100.
• Solution:Taking u = g(x) = x3 – 1 and n = 100
= (x3 – 1)100
= 100(x3 – 1)99 (x3 – 1)
= 100(x3 – 1)99 3x2
= 300x2(x3 – 1)99
Implicit Differentiation
So far we worked with functions where one variable is expressed in terms of another variable—for example:y = or y = x sin x (in general: y = f (x). )
Some functions, however, are defined implicitly by a relation between x and y, examples:
x2 + y2 = 25x3 + y3 = 6xy
We say that f is a function defined implicitly - For example Equation 2 above means: x3 + [f (x)]3 = 6x f (x)