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More on Derivatives and Integrals -Product Rule -Chain Rule AP Physics C Mrs. Coyle

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f ’ (x) = lim f( x + h) f( x ) h 0 h Derivative

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Derivative Notations f ’ (x) df (x) dx. f df dx

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Notations when evaluating the derivative at x=a f(a) df (a) dx f’(a) df | x=a dx

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Basic Derivatives d(c) = 0 dx d(mx+b) = m dx d(x n ) = n x n-1 dx n is any integer x≠0

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Derivative of a polynomial. For y(x) = ax n dy = a n x n-1 dx - Apply to each term of the polynomial. -Note that the derivative of constant is 0.

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Product Rule For two functions of x: u(x) and v (x) d [u(x) v (x)] =u d v (x) + v d u (x) dx dx dx or (uv)’ = u v’ + vu’

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Example of Product Rule: Differentiate: F=(3x-2)(x 2 + 5x + 1) Answer: F’(x) = 9x 2 + 26x-7

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Chain Rule If y=f(u) and u=g(x): dy = dy du dx du dx

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Example of Chain Rule Differentiate: F(x)= (x 2 + 1) 3 Ans:F’(x)= 6(x 2 +1) 2 x

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Second Derivative Notations df ’ (x) dx d 2 f (x) d x 2 f’’(x)

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Example of Second Derivative Compute the second derivative of y=(x) 1/2 Ans: (-1/4) x -3/2

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Derivatives of Trig Functions d sinx = cosx dx d cosx = -sinx dx d tanx = sec 2 x dx d secx = secx tanx dx

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Derivative of the Exponential Function d e u = e u du dx dx

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Example of derivative of Exponential Function 2 Differentiate: e x 2 Ans: 2x e x

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Derivative of Ln d (lnx) = 1/x dx

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Definite Integral b a ∫ b f(x) dx= F(b)-F(a)= F(x)| a a and b are the limits of integration.

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If F(x)= ∫ f(x) dx then d F(x) = f(x) dx

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Properties of Integrals a ∫ c f(x) dx = a ∫ b f(x) dx+ b ∫ c f(x) dx a**
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Basic Integrals (integration constant ommited) ∫ x n dx = 1 x n+1, n ≠ 1 n+1 ∫ e x dx = e x ∫ (1/x) dx = ln|x| ∫ cosx dx = sinx ∫ (1/x) dx = ln|x| ∫ sinx dx = -cosx

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Example with computing work. There is a force of 5x 2 –x +2 N pulling on an object. Compute the work done in moving it from x=1m to x=4m. Ans: 103.5N

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To evaluate integrals of products of functions : Chain Rule Integration by parts Change of Variable Formula

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When a function and its derivative appear in the integral: a ∫ b f[g(x)]g’(x) dx = g(a) ∫ g(b) f(y) dy

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Example: When a function and its derivative appear in the integral: Compute x=0 ∫ x=1 2x (x 2 +1) 3 dx Ans: 3.75 Ans:

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Example of Change of Variable Formula Evaluate: 0 ∫ 1 2x (x 2 + 1) 9 dx Answ: 102.3

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Integration by Parts a ∫ b u(x) dv dx= dx b = u(x) v(x)| a - a ∫ b v(x) du dx dx

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Integration by Parts b a ∫ b u v’ dx= u v| a - a ∫ b v u’ dx

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Example of Integration by Parts Compute 0 ∫ π x sinx dx Ans: π

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