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More on Derivatives and Integrals -Product Rule -Chain Rule. AP Physics C Mrs. Coyle. Derivative. f ’ (x) = lim f( x + h) - f( x ) h 0 h. Derivative Notations. f ’ (x) df (x) dx. . f df dx. Notations when evaluating the derivative at x=a. f(a) df (a) dx. f’(a)
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More on Derivatives and Integrals-Product Rule-Chain Rule AP Physics C Mrs. Coyle
Derivative f’ (x) = limf(x + h) - f(x ) h0h
Derivative Notations f’ (x) df (x) dx . f df dx
Notations when evaluating the derivative at x=a f(a) df (a) dx f’(a) df |x=a dx
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
Derivative of a polynomial. For y(x) = axn dy = a n xn-1 dx -Apply to each term of the polynomial. -Note that the derivative of constant is 0.
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) dxdxdx or (uv)’ = u v’ + vu’
Example of Product Rule: Differentiate: F=(3x-2)(x2 + 5x + 1) Answer: F’(x) = 9x2 + 26x-7
Chain Rule If y=f(u) and u=g(x): dy = dydu dx du dx
Example of Chain Rule Differentiate: F(x)= (x 2 + 1) 3 Ans:F’(x)= 6(x2 +1)2x
Second Derivative Notations df’ (x) dx d2f (x) dx2 f’’(x)
Example of Second Derivative Compute the second derivative of y=(x)1/2 Ans: (-1/4) x-3/2
Derivatives of Trig Functions dsinx = cosx dx dcosx = -sinx dx dtanx = sec2 x dx dsecx = secxtanx dx
Derivative of the Exponential Function d e u = e u du dxdx
Example of derivative of Exponential Function 2 Differentiate: ex 2 Ans: 2x e x
Derivative of Ln d (lnx) = 1/x dx
Definite Integral b a∫b f(x) dx= F(b)-F(a)= F(x)|a a and b are the limits of integration.
If F(x)=∫ f(x) dx then d F(x) = f(x) dx
Properties of Integrals a∫c f(x) dx =a∫b f(x) dx+b∫c f(x) dx a<b<c a∫bcf(x) dx =ca∫b f(x) dx • a∫b (f(x)+g(x)) dx =a∫bf(x) dx+ a∫b g(x) dx
Basic Integrals (integration constant ommited) ∫ xndx = 1 xn+1 , n ≠ 1 n+1 ∫ exdx = ex ∫ (1/x) dx = ln|x| ∫ cosxdx = sinx ∫ sinxdx = -cosx ∫ (1/x) dx = ln|x|
Example with computing work. • There is a force of 5x2 –x +2 N pulling on an object. Compute the work done in moving it from x=1m to x=4m. • Ans: 103.5N
To evaluate integrals of products of functions : • Chain Rule • Integration by parts • Change of Variable Formula
Change of Variable Formula 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
Example: When a function and its derivative appear in the integral: • Compute x=0∫x=1 2x (x2 +1) 3 dx • Ans: 3.75 • Ans:
Example of Change of Variable Formula Evaluate: 0∫1 2x (x2 + 1) 9 dx Answ: 102.3
Integration by Parts a∫b u(x) dvdx= dx b = u(x) v(x)|a - a∫bv(x) du dx dx
Integration by Parts b a∫b u v’ dx= u v|a - a∫bv u’ dx
Example of Integration by Parts Compute 0∫π x sinx dx Ans: π