how to find dy dx
Derivatives as dy/dx
Derivatives are all about change ...
... they show how fast something is changing (called the rate of change) at any point.
In Introduction to Derivatives (please read it first!) we looked at how to do a derivative using differences and limits.
Here we look at doing the same thing but using the "dy/dx" notation (also called Leibniz's notation) instead of limits.
We start by calling the function "y":
y = f(x)
1. Add Δx
When x increases by Δx, then y increases by Δy :
y + Δy = f(x + Δx)
2. Subtract the Two Formulas
From: | y + Δy = f(x + Δx) | |
Subtract: | y = f(x) | |
To Get: | y + Δy − y = f(x + Δx) − f(x) | |
Simplify: | Δy = f(x + Δx) − f(x) |
3. Rate of Change
To work out how fast (called the rate of change) we divide by Δx:
Δy Δx = f(x + Δx) − f(x) Δx
4. Reduce Δx close to 0
We can't let Δx become 0 (because that would be dividing by 0), but we can make it head towards zero and call it "dx":
Δx dx
You can also think of "dx" as being infinitesimal, or infinitely small.
Likewise Δy becomes very small and we call it "dy", to give us:
dy dx = f(x + dx) − f(x) dx
Try It On A Function
Let's try f(x) = x2
dy dx | = f(x + dx) − f(x) dx | ||
= (x + dx)2 − x2 dx | f(x) = x2 | ||
= x2 + 2x(dx) + (dx)2 − x2 dx | Expand (x+dx)2 | ||
= 2x(dx) + (dx)2 dx | x2−x2=0 | ||
= 2x + dx | Simplify fraction | ||
= 2x | dx goes towards 0 |
So the derivative of x2 is 2x
Why don't you try it on f(x) = x3 ?
dy dx | = f(x + dx) − f(x) dx | ||
= (x + dx)3 − x3 dx | f(x) = x3 | ||
= x3 + ... (your turn!) dx | Expand (x+dx)3 |
What derivative do you get?
how to find dy dx
Source: https://www.mathsisfun.com/calculus/derivatives-dy-dx.html
Posted by: reynoldsbuthadou.blogspot.com
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