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Criswell Calculus
AP Calculus Learning to Take Derivatives by Rules (Chapter 3)
| Question | Answer |
|---|---|
| Definition: The Intermediate Value Theorem (IVT) | If "f" is a continuous function on the closed interval [a,b] and K falls between f(a) and f(b), then there must exist at least one c on the open interval (a,b) where f(c) =k |
| Definition: Average Rate of Change | As a formula: {F(b) –F(a)} / (b –a), which would be the same as the definition of slope m = (y2 –y1) / (x2 – x1). |
| Definition: Derivative of a function | "Global Derivative" Lim h–>0 [ f (x+h) – f(x) ] / h Be able to extract from the notation, "What is the function?" and at "What x value?" |
| What does it mean to take the derivative of a function? | Instantaneous Rate of Change. Slope of the tangent line for any point on the curve. |
| Relationship of Derivatives on Projectiles | Displacement or Position Function: s(t), Velocity: s '(t) = v(t), Acceleration: s '' (t) = v ' (t) = a(t). |
| Derivative at a Point | "Disposable Derivative". Lim x–>c [ f (x) – f(c) ] / (x – c) Be able to extract from the notation, "What is the function?" and at "What x value?" |
| Mean Value Theorem: Finding x -value where m-tangent = m-secant | States that if a function f is continuous on the closed interval [a,b], and differentiable on the open interval (a,b), then there exists at least point c in the interval (a,b) such that f'(c) is equal to the average rate of change over [a,b]. |
| Estimating the Instantaneous Rate of Change | The best estimate from a table of values will come from using values that are a little under and a little over the desired target. Example f' (5) ≈ m sec =[f (5.1) – f (4.9) ] / (5.1 –4.9) even if f(5) is known it should NOT be used in the estimate! |
| Power rule for exponents | If f(x) = a*x^n, then f ' (x) = n*a*x^(n-1) This concept can be extended independently across numerous terms in an equation . |
| Product Rule | if the function can be split into a product such as y = u*v, then y ' = u*d(v) + v*d(u) Be prepared for a play on notation such as h(x)=f(x)*g(x) whereas h'(x) = f(x)* g '(x) + g(x)* f ' (x) |
| d(sin (x)) / dx | cos (x) |
| d(cos (z)) / dz | - sin (z) |
| d(tan (b)) / db | sec^2(b) |
| d(csc (r)) / dr | - csc(r)*cot(r) |
| d(sec (w)) / dw | sec(w)*tan(w) |
| d(cot(g))/ dg | - csc^2 (g) |
| Dx ( ln(x) ) | 1/x |
| Dx ( e^x ) | e^x |
| Understand that ln (x) and e^x are inverse functions. | if y = ln | e^(3x-7) |, then an equivalent form y = 3x -7 if f(x) = e^(ln |2x+5|), then an equivalent form f(x) = 2x+5 |
| Quotient Rule | Generically d(high /low) = [low*d(high) - high*d(low) ] / low^2 Again be prepared for notation questions: if h(x) = f(x) / g(x) , then h '(x) = [g(x)*f '(x) - f(x)* g '(x)] / ( g(x) )^2 |
| Be able to confirm Dx (tan(x)) and Dk (cot(k)) using the quotient rule. | Example: if f (x) = tan(x) equivalent form f(x) = (sin(x) / cos(x)) so, f' (x) = [ cos(x)*d(sin(x)) - sin(x)*d(cos(x)) ] / (cos(x))^2 = [cos^2(x) + sin^2(x)]/ (cos(x))^2 =1 / (cos(x))^2 =sec^2 (x) |
| Mean Value Theorem (MVT) and Desmos | When working with the MVT on edia if you run into a function that you are not 100% certain on how to differentiate, there is a calculus function option that will graph the derivative to a function on Desmos. This could be extremely helpful. |
Created by:
Troy.Criswell
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