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2.2 Basic Differentiation Rules Find the derivative of a function using the constant rule and power rule. Find the derivatives of the sine function and of the cosine function. Use derivatives to find rate of change.
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Constant Rule The derivative of a constant function is 0. That is, if c is a real number, then d/dx[c] = 0. 1) f(x) = 22) y = -⅖ Why?
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Power rule If n is a rational number, then the function f(x) = xⁿ is differentiable and d/dx [xⁿ] = n xⁿ⁻¹ 1) f(x) = x³2) y = x3) g(x) = x⁻⁷
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Graphically, what does a derivative of a polynomial function look like?
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Constant Multiple Rule If f is a differentiable function and c is a real number, then cf is also differentiable and d/dx [cf(x)] = c(f´(x)) This is a combination of the Power & Constant Multiple d/dx [cxⁿ] = cnxⁿ⁻¹ 1) -3x³2) ¼x⁻⁴ Often times functions have to be rewritten to use the power rule.
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Sum and Difference Rules The sum ( or difference) of two differentiable functions f and g is itself differentiable. Then f + g or f – g is the derivatives of f and g. d/dx [f(x) + g(x)] = f’(x) + g’(x) d/dx [f(x) - g(x)] = f’(x) - g’(x)
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Derivatives of Sine and Cosine functions d/dx [sin(x)] = cos(x) d/dx [cos(x)] = - sin(x) Proof uses the definition of derivative and trigonometry identities.
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Rate of change f’(x) = m = rate of change Applications involving rates of change occur in a wide variety of fields. Like … Often time we will discuss the rate of change as it refers to the motion of a object in relation to time, this is called a position function s(t).
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Average velocity is Now as the change in time approaches 0 ( i.e. the limit as t→0) you can find the instantaneous velocity. Which means we can find the velocity of an object at any specific time(in the domain) not just over an interval of time.
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Therefore, the derivative of the position function is equal to the velocity function. s’(t) = v(t) Negative v(t) Positive v(t) Zero v(t) Speed of an object is the absolute value of velocity. Accleration
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The position of a free-falling object under the influence of gravity can be represented by the equation s(t) = ½gt² + v₀t + s₀ Where s₀ is the initial height of the object, v₀ is the initial velocity of the object, and g is the acceleration due to gravity. On Earth gravity is -32ft/sec² OR -9.8 m/sec²
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Example At time t=0, a diver jumps form a platform diving board that is 32 feet above the water. The position of the diver is s(t) = -16t² + 16t + 32 where s is in feet and t is in seconds. a)When does the diver hit the water? b)What is the diver’s velocity at impact? c)At what time did the diver reach maximum height? d)When was the diver’s velocity positive, zero and negative?
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