Overview of topic:
-Differentiation: a technique for measuring the slope at a curve at any point.
-Newton and Leibniz in the seventeenth century not only defined the slope of the curve at any point as being the slope of the tangent to the curve at that point, but also showed how the slope could be calculated.
-The technique of differentiation can be used for finding marginal cost and marginal revenue, demand and supply elasticities, and a firm’s most profitable output.
Slope/Gradient/Rate of change = distance travelled up RISE / RUN distance travelled forward
/The Difference Quotient = y1-y0 / x1 – x0
= Δy / Δx
= increase in y / increase in x
The following:
Δy / Δx = y1-y0 / x1 – x0 = a(x1 – x0) / x1 – x0) = a
Tells us:
- The slope of the linear function y=ax+b is given by the parameter a.
- The slope equals a, wherever we measure it. Since a is a constant, the slope is a constant.
Progress Exercises 6.1:
- For each of the following linear functions, calculate the difference quotient for the specified change in x. In each case, sketch the graph and indicate where you have measured the difference quotient.
- y=2x+4 x0=2 x1=3
- y=½x-2 x0=-1 x1=-½
- y=50x x0=1 x1=10
Answers
- (2×1+4) – (2×0+4) / x1-x0 = 2(x1-x0)/x1-x0 = 2
- (½x1-2) – (½x0-2) / x1-x0 = ½(x1-x0)/x1-x0 = ½
- 50×1 – 50×0 / x1-x0 = 50(x1-x0)/x1-x0 =50
The tangent as a measure of slope: A tangent to a curve is defined as a straight line that touches the curve at only one point.
Step 1: Finding the difference quotient
Step 2: Using the difference quotient to find the slope of the tangent
However, both these steps are long and tedious, therefore we can use the following rules instead:
Power rule: y=x^n the derivative is: dy/dx = nxᶯ-¹
E.g. x² = 2x²-1, so dy/dx = 2x
E.g. y=x = n=1, so dy/dx = 1x^1-1 = x^0 = 1
Multiplicative constant: A f(x) = dy/dx = A f’’(x)
E.g. y=10x² = 20x
E.g. y=15x³ = 45x²
E.g. y = 1/50x^½ = 1/100x^-½
Additive constant: y = f(x) + B, dy/dx = f’(x)
E.g. y = x²+62, dy/dx = f’(x) = 2x
Sums or Differences: if y=f(x)+g(x), then dy/dx = f’(x) + g’(x)
E.g. y = 10x²+x^½ , dy/dx = 20x+1/2x^-½
E.g. y= 10x²-x^½, dy/dx = 20x-1/2x^-½
E.g. y=x³+5x²-3x, dy/dx = 3x²+10x-3
Progress Exercises 6.2:
- y=5x² =10x
- y=x^4 =4x³
- y=12x³ =36x²
- y=x^0.5 =0.5x^-0.5
- y=x³+x² =3x²+2x
- y=3x^4 +10x² =12x³+20x
- y=4x³-16x+9 =12x²-16
- y=10^3+5x²-9x+5/x =30x²+10x-9-5x^-2
- y=⅕x³+1/2x²+5x+3 =3/5x²+x+5
- y=x^1/2+x^-2 =1/2x^-½+2x^-3
- y=x^½-x^-2 =1/2x^-½+2x^-3
- y=-2x^-2+x^-0.5+x =4x^-3-0.5x^-0.5+1
- y=1+1/x-x^-0.25 =-x^-2+0.25x^-1.25
- y=x+(1/x²) =1-2x^-3 (since 1/x² = x^-2)
- y=-2x^5-0.3x^-0.3+1/2x^½ =-10x^4+0.9x^-01.3+1/4x^-1/2
- y=ax+b (where a and b are parameters) =a
- y=ax² +bx +c (where a, b, c are parameters) =2ax+b
- q=2p³+5p² (p and q instead of x and y) =6p²+10p
- q=Ap^-a (where q and p are variables, A and a are parameters) =-aAp^-a-1
- z=0.5t³ + 4t+15 =1.5t²+4
Function of a function rule (Chain Rule): dy/dx = dy/du du/dx
E.g. y=(x²+5x)³, dy/dx = 3(x²+5x)²(2x+5)
E.g. y=(x²+1)^5, dy/dx = 5(x²+1)^4(2x) = 10x(x²+1)^4
Product Rule: y=uv, dy/dx = u dv/dx + v du/dx
E.g. y= (x²+1)(x³+x²), dy/dx = u dv/dx + v du/dx = (x²+1)(3x²+2x)+(x³+x²)(2x) = 5x^4+4x³+3x²+2x
E.g. y=(5x²+3x)(x^4+x), (5x²+3x)(4x³+1)+(x^4+x)(10x+3) = 30x^5+15x^4+15x²+6x
Quotient Rule: dy/dx = [v du/dx – u dv/dx] / v²
E.g. y=x²+1/x³+x² = (x³+x²)(2x)-(x²+1)(3x²+2x) / (x³+x²)² = dy/dx = -x³-3x-2/x³(x+1)²
E.g. y=2x+1/3x²+3x+1 = (3x²+3x+1)(2)-(2x+1)(6x+3) / (3x²+3x+1)² dy/dx = -6x²-6x-1/(3x²+3x+1)²
The Derivative of the Inverse Function: y=f(x), dy/dx= 1/dx/dy
E.g. y=3x+2, dx/dy = x=1/3y-2/3
E.g. x=y² = 2y, 1/2y
Progress Exercise 6.3:
From rules 5-8:
- y=(1+2x)² = 4(1+2x)
- y=(1-x²)^½ = -x(1-x²)^-1/2
- y=(x²-2x³)^5 = 5(x²-2x³)^4(2x-6x²)
- y=(x²-x)^0.5 =now you calculate…
- y=(3x-2)(2x+1) =
- y=(4x²-3)(2x^5+x) =
- y=x²+1 / 1-x² =
- y=(x²+1)^½ / 1-x² =
- y=(y+1)^0.1 =
- y=1/1-x =
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