First Derivative Equations

Question 1

Power Rule

Answer 1

d/dx[x^n] = nx^n-1

Question 2

Sin/Cos/Tan Rule

Answer 2

d/dx[sin x] = cos x
d/dx[cos x] = -sin x
tan x = sin x / cos x
tan’ x = (sin x)^2 + (cos)^2 / (cos x)^2 = sec^2x

Question 3

Exponential Rule

Answer 3

d/dx[e^u] = e^u . u’

Question 4

Product Rule

Answer 4

d/dx[x . y] = x’y + xy’

Question 5

Chain Rule/Composition Rule

Answer 5

d/dx[F[g(x)]) = f’[g(x)] . g’(x) where F is a power
Implementing power rule:
f’[g(x)] = f * [g(x)]^f-1

Question 6

Differentiation

Answer 6

A type of equation which allows us to find if the equation has a critical point, aka if the first derivative is equal to 0:
y2 - y1 / x2 - x1 = f(t+h) - f(t) / t+h - t (with h being the difference between y2 and y1 etc).

Question 7

Second Derivative Test

Answer 7

1) Construct f’(x).
2) Find all solution, a, for which f’(a) = 0.
3) Construct f’‘(x), the second derivative.
4) For each of solution of a in step 2, we find the derivatives of f’‘(a). If f’‘(a) < 0, then a is a local maximum. If f’‘(a) > 0, then its a local minimum. If it equals 0, then no conclusion can be made.
Example:
f(x) = x^4 - 3x^2 + 1
Applying the power rule we have x^4 = 4x^3, and 1 = 0.
Applying the product rule to -3x^2, since it is -3 multiplied by x^2 gives us -6x.
The first derivative turns out to be 4x^3 - 6x.
Three critical points are 0, + √3/2 and √-3/2.
f’‘(x) = 12x^2 - 6.
f’‘(0) < 0 since it equals -6, giving us the local maximum.
f’’(√3/2) = 12, and f’’(-√3/2) = 12, giving us our local minima’s.

Question 8

Second Derivative Test With 2-Variable Functions

Answer 8

We first take the derivative of the function twice, one in respect to x and y is fixed; fx(x, y), and then one in respect to y and x is fixed; fy(x, y).

We can use the second derivative test on these two variables to conjure 4 forms; fxx, fyy, fxy, fyx, for example, taking fx(x, y) and then apply the derivative of that answer in respect to y whilst x is fixed, resulting in fxy(x, y).