Derivatives



Derivatives

|Function |Derivative |

|xn |nx n-1 |

|ex |ex |

|ax |(ln a)ax |

|uv |[pic] |

|[pic] |[pic] |

|If y = f(u) and |[pic] |

|u = g(x), then | |

|sin x |cos x |

|cos x |- sin x |

|tan x |[pic] |

|ln x * |[pic] |

|arctan x * |[pic] |

|arcsin x * |[pic] |

*Derived on pgs. 226-27 of Hughes-Hallett

Integrals

|Function |Integral |

|xn |[pic] |

|ex |ex |

|sin x |- cos x |

|cos x |sin x |

|tan x |- ln | cos x | |

|[pic] |tan x |

|[pic] |ln x |

|[pic] |[pic] |

|ln x |x ln x – x , x > 0 |

|ax |[pic] |

Implicit Differentiation

To implicitly differentiate an expression of the form, (say),

[pic], we take [pic]

Where [pic] and we use the product rule for [pic] to get:

[pic]

We then “solve for” [pic] :

[pic]

[pic]

[pic]

[pic] (Then one can calculate the value of the derivative at a given point by plugging

in the x and y values at that point.)

We may also apply the chain rule to implicit differentiation:

Ex. sin(xy) = 2x + 5

[pic]

Note that y = sin u and u = xy, so [pic], [pic],

so [pic]. Returning to the above implicit differentiation:

[pic]

[pic]

Integration Methods

1. Substitution

Let w be an “inside function” whose derivative, [pic], will “cancel out” other terms. We then reassemble an integral of the form [pic] into the form [pic].

Ex. [pic] Let w = x3 , then [pic] and dw = 3x2dx

[pic]

2. Integration by Parts (Formula is shown in anti-derivative table)

Ex. [pic] u = ln x [pic]

[pic] [pic]

[pic] [pic]

Note: Often when an integral is in the form ∫ xa f(x) dx , it is useful to let u = xa. (Certainly not always, as shown in the example above)

3. Partial Fractions

Ex. [pic]

[pic]

1 = A(x – 5) + B(x - 2)

Let x = 5, then 1 = 3B , so B = 1/3

Let x = 2, then 1 = - 3A, so A = - 1/3.

So, [pic]

Note: If the highest power of the term in the numerator is greater than or equal to the term in the denominator, you will likely need to perform a division, get a remainder, and express the fraction as a non-fraction plus a fractional denominator.

4. Estimating the Integral Function when the Anti-Derivative Cannot be Found Algebraically

If an integral [ F(x) ] cannot be found algebraically, we can still find values of the integral function by estimating the area under the curve of f(x). [where F′(x) = f(x) ].

Ex. [pic]

|x |0 |1 |2 |3 |

|F(x) |0 |0.95 |1.61 |1.85 |

Ex. If the function g defined by [pic] on the closed interval - 1 ≤ x ≤ 3, then g has a local minimum at x = ?

[pic]

Note that for the domain - 1 ≤ x ≤ 3, the area is positive between – 1 and 1.8, then some area is “subtracted off”. The location where the most area has been “subtracted off” corresponds to the local minimum of g.

Likewise, we can see that the maximum value of g on the interval - 1 ≤ x ≤ 3 must occur at approximately x = 1.8.

Ex. If [pic] and f(0) = 1, then what is f(2)?

In this type of estimation, we normally assume that the y-intercept occurs at the origin. Since f(0) = 1 in this case, we know that all values of f have been “pushed up” by 1.

So, in calculating the area of [pic] on the calculator from 2 to 0, the calculator gives the value 0.16, so the solution for f(2) = 0.16 + 1 ≈ 1.16.

Solving First-Order Differential Equations

This will be an equation in which we are shown the derivative function, and be asked to determine the initial function. Obviously, the solution to any given differential equation will be a whole family of functions (each with a different constant), but if we are given some initial condition, we can solve for the particular equation.

To solve this, we simply “gather” the y’s on one side and the x’s on the other, then integrate both sides. Often, we will want to then isolate the y.

Ex. [pic] [pic] [pic] [pic]

[pic] [pic], but eC is just some constant, so let A = eC ,

then, [pic]

Ex. [pic]Determine the solution to the differential equation [pic] if y(2) = 3.

[pic] [pic] [pic] , but when x = 2, y = 3, so

[pic] 9 = 4 + C , so C = 5:

[pic] [pic] [pic]

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