 More on Derivatives and Integrals -Product Rule -Chain Rule

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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) 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) dx dx dx or (uv)’ = u v’ + vu’

Example of Product Rule:
Differentiate: F=(3x-2)(x2 + 5x + 1) Answer: F’(x) = 9x2 + 26x-7

If y=f(u) and u=g(x): dy = dy du dx du dx
Chain Rule If y=f(u) and u=g(x): dy = dy du 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) d x2 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
d sinx = cosx dx d cosx = -sinx d tanx = sec2 x dx d secx = secx tanx

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

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

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

a∫b f(x) dx= F(b)-F(a)= F(x)|a
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∫b cf(x) dx =c a∫b f(x) dx a∫c f(x) dx = a∫b f(x) dx+ b∫c f(x) dx a<b<c a∫b (f(x)+g(x)) dx = a∫b f(x) dx+ a∫b g(x) dx

Basic Integrals (integration constant ommited)
∫ xn dx = 1 xn+1 , n ≠ 1 n+1 ∫ ex dx = ex ∫ (1/x) dx = ln|x| ∫ cosx dx = sinx ∫ sinx dx = -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) dv dx= dx b = u(x) v(x)|a - a∫b v(x) du dx dx

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

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