Calculus

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Differentiating from first principles;

Consider y = x^2;

The gradient at A[2 , 4] is given by the tangent A.

You can find this value by differentiating from first principles.

An approximation to the tangent is the Chord AP, where P is a point on y = x^2. The x-coordinate of P is 2 + h, as h becomes smaller the chord AP gets closer to the tangent [2 , 4].

P's Y coordinate is; (x + h)^2 = 4 + 4h + h^2

P; [2 + h , 4 + 4h + h^2]

So to find the gradient of AP use;

[x1 , y1] to [x2 , y2] = (y2 - y1)/(x2 - x1)

So as h --> 0

The gradient of chord AP --> the gradient of tangent at A

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Cubic and Quartic Functions;

Need the expansions of (a + h)^3 and (a + h)^4

(a + h)^3 => (a + h)(a + h)(a + h)

=> (a^2 + 2ah + h^2)(a + h)

=> (a^3 + 2a^2h + ah^2 + ha^2 + 2ah^2 + h^3)

=> (a^3 + h^3 + ah(3a + 3h))

(a + h)^4 => (a + h)(a + h)(a + h)(a + h)

=> (a^2 + 2ah + h^2)(a^2 + 2ah + h^2)

=> (a^4 + 2a^3h + a^2h^2 + 2a^3h + 4a^2h^2 + 2ah^3 + a^2h^2 + 2ah^3 + h^4)

=> (a^4 + h^4 + 6a^2h^2 + 4a^3h + 4ah^3)

=> (a^4 + h^4 + 2ah(3ah + 2a^2 + 2h^2))

Página 3

Improper Integrals;

If the integral has either ∞ or -∞ as one of it's limits then they are improper integrals.

The intergrand is undefined at its limits for integration; Therefore the function may have a denominator which becomes 0 for a value of x

Example 1;

Determine ∫ (limits ∞ ---> 1) 1/(x^2) dx

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∫ (limits ∞ ---> 1) 1/(x^2) dx = ∫ (limits n ---> 1) (x)^-2 dx

= [((x)^-1)/-1)] [Limits n ---> 1]

= [(-1/x)] [Limits n ---> 1]

= (-(1/n) + 1)

n --> ∞ the value approaches. Therefore, the improper integral ∫ (limits ∞ ---> 1) 1/(x^2) dx and is = 1

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