1. ## Simple Derivatives

Hello,

I'm teaching myself Calculus, as I see a need for it in my future career. I started watching some basic calculus videos from MIT, but there are a few questions I would like to ask:

I am aware of the following rule:

$\displaystyle \frac{dy}{dx} = nx^{n-1}$

However, I don't understand how to apply it to a function such as this:

$\displaystyle y = 1 - x^2$

Or this:

$\displaystyle y = (x^3 - 2x^2)$

I would appreciate any assistance!

Thanks,

Austin

2. ## Re: Simple Derivatives

Originally Posted by ElectroNerd
Hello,

I'm teaching myself Calculus, as I see a need for it in my future career. I started watching some basic calculus videos from MIT, but there are a few questions I would like to ask:

I am aware of the following rule:

$\displaystyle \frac{dy}{dx} = nx^{n-1}$

However, I don't understand how to apply it to a function such as this:

$\displaystyle y = 1 - x^2$

Or this:

$\displaystyle y = (x^3 - 2x^2)$

I would appreciate any assistance!

Thanks,

Austin
you also need to be aware that the derivative (rate of change) of a constant is 0

$\displaystyle y = 1 - x^2$

$\displaystyle \frac{dy}{dx} = 0 - 2x = -2x$

for the second, just use the power rule ...

$\displaystyle y = x^3 - 2x^2$

$\displaystyle \frac{dy}{dx} = 3x^2 - 4x$

3. ## Re: Simple Derivatives

You also need to know that $\displaystyle \frac{d(f+ g)}{dx}= \frac{df}{dx}+ \frac{dg}{dx}$ and $\displaystyle \frac{d(cf)}{dx}= c \frac{df}{dx}$ where c is any constant.
(You typically learn those before you learn $\displaystyle \frac{d(x^n)}{dx}= nx^{n-1}$.)

$\displaystyle \frac{d(1- x^2)}{dx}= \frac{d(1)}{dx}+ \frac{d(-x^2)}{dx}= \frac{(x^0)}{dx}- \frac{d(x^2)}{dx}$ where I have used the fact that $\displaystyle x^0= 1$ for all x except 0.

$\displaystyle \frac{d(x^3- 2x^2)}{dx}= \frac{d(x^3)}{dx}+ \frac{d(-2x^2)}{dx}= \frac{d(x^3)}{dx}- 2\frac{d(x^2)}{dx}$