Calculus

Question 1

Computing Gradients

Answer 1

We take the difference between y1 and y2, and difference between x1 and x2.
So the equation for the gradient would be:
y2 - y1 / x2 - x1

Question 2

The Gradient of Importance

Answer 2

The gradient of importance when analysing a function f(x) are those of “the line touching the curve (x, f(x))”, not the gradient of the tangent.
Such gradients are described as the first derivative of f(x), denoted as f’(x).

Question 3

Equation For The Closest Point’s Gradient

Answer 3

f(t+h) - f(t) / (t+h) - t

f(t+h) - f(t) / h

With h being the distance from the original point.

This is because the original point is equal to (f(x), x).

Question 4

First Derivative

Answer 4

Lets say f(x) = (25 - x^2) / (5-x). We can use this to find the first derivative:
(5+x)(5-x) / (5-x) = (5+x), therefore f(x) = 5+x.
Therefore it cannot be 0, since 1 is the first derivative (via differentiation).
A more thorough explanation:
f’(x) = f(x+h) - f(x) / h
f(x+h) = 5+(x + h), since we now add h to x.
Therefore f’(x) = (5 + (x+h)) - (5 + x) / h, the equation for computing gradients.
That becomes 5 + x + h - 5 - x / h.
That becomes h, leaving us with h/h.
h/h = 1 since a number divided by itself is equal to one.