DT.01.1 - Derivative Rules - Power Rule, Constant Rule, Sum and Difference Rule

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DT.01.1 - Derivative Rules - Power Rule, Constant Rule, Sum and Difference Rule

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DT.01.1 - Derivative Rules - Power Rule, Constant Rule, Sum and Difference Rule

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DT.01.1 - Derivative Rules - Power Rule, Constant Rule, Sum and Difference Rule

MCB4U - Santowski

- The equation used to find a tangent line or an instantaneous rate of change is:
- which we also then called a derivative.
- So derivatives are calculated as .
- Since we can differentiate at any point on a function, we can also differentiate at every point on a function (subject to continuity) therefore, the derivative can also be understood to be a function itself.

- We will now develop a variety of useful differentiation rules that will allow us to calculate equations of derivative functions much more quickly (compared to using limit calculations each time)
- First, we will work with simple power functions
- We shall investigate the derivative rules by means of the following algebraic and GC investigation (rather than a purely “algebraic” proof)

(i) f(x) = 3 is called a constant function graph and see why.

What would be the rate of change of this function at x = 6? x = -1, x = a?

We could do a limit calculation to find the derivative value but we will graph it on the GC and graph its derivative.

So the derivative function equation is f `(x) = 0

(ii) f(x) = 4x or f(x) = -½x or f(x) = mx linear fcns

What would be the rate of change of this function at x = 6? x = -1, x = a.

We could do a limit calculation to find the derivative value but we will graph it on the GC and graph its derivative.

So the derivative function equation is the constant function f `(x) = m

(iii) f(x) = x2 or 3x2 or ax2 quadratics

What would be the rate of change of this function at x = 6? x = -1, x = a.

We could do a limit calculation to find the derivative value but we will graph it on the GC and graph its derivative.

So the derivative function equation is the linear function f `(x) = 2ax

(iv) f(x) = x3 or ¼ x3 or ax3?

What would be the rate of change of this function at x = 6? x = -1, x = a.

So the derivative function equation is the quadratic function f `(x) = 3ax2

(v) f(x) = x4 or ¼ x4 or ax4?

What would be the rate of change of this function at x = 6? x = -1, x = a.

So the derivative function equation is the cubic function f `(x) = 4ax3

- We can summarize the observations from previous slides the derivative function is one power less and has the coefficient that is the same as the power of the original function
- i.e. x4 4x3
- GENERAL PREDICTION xn nxn-1

- do a limit calculation on the following functions to find the derivative functions:
- (i) f(x) = k
- (ii) f(x) = mx
- (iii) f(x) = x2
- (iv) f(x) = x3
- (v) f(x) = x4
- (vi) f(x) = xn

- We can summarize our findings in a “rule” we will call it the power rule as it applies to power functions
- If f(x) = xn, then the derivative function f `(x) = nxn-1
- Which will hold true for all n
- (Realize that in our investigation, we simply “tested” positive integers we did not test negative numbers, nor did we test fractions, nor roots)

- We have seen derivatives of simple power functions, but what about combinations of these functions? What is true about their derivatives?

(vi) f(x) = x² + 4x - 1

What would be the rate of change of this function at x = 6? x = -1, x = a.

So the derivative function equation is the “combined derivative” of f `(x) = 2x + 4

(vii) f(x) = x4 – 3x3 + x2 -½x + 3

What would be the rate of change of this function at x = 6? x = -1, x = a.

So the derivative function equation is the “combined derivative” f `(x) = 4x3 – 3x2 + 2x – ½

- Visual Calculus - Differentiation Formulas
- Calculus I (Math 2413) - Derivatives - Differentiation Formulas from Paul Dawkins
- Calc101.com Automatic Calculus featuring a Differentiation Calculator
- Some on-line questions with hints and solutions

- MCB4U – Nelson text,
- page 222, Q2efghij,3efghi,5cdehi,7,8,16,13
- page 229, Q2,3,9,11,13,14
- IBHLY2 – Stewart, 1989, Chap 2.2, p83, Q7-10
- Stewart, 1989, Chap 2.3, p88, Q1-3eol,6-10