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Posted: **August 24th, 2011, 1:04 pm**

The derivative of

with respect to x is the function

and is defined as,

Example

First plug the function into the definition of the derivative.

Multiply everything out

So the derivative is

Posted: **October 18th, 2011, 10:08 pm**

this is the alternate definition of the derivative.

$\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$

Posted: **October 23rd, 2011, 9:01 pm**

Rules

These will make everything much easier

Rule 1
Rule 2
Rule 3
Rule 4
Rule 5
Horizontal Tangents
Rule 6

Posted: **October 23rd, 2011, 9:02 pm**

Rule 1: Derivative of a Constant

If c is a constant, then $\frac{d}{dx}(c)=0$

Rule 2: Power Rule for a Positive Integer Powers of x

If n is a positive integer, then $\frac{d}{dx}(x^n)=$ nxⁿ⁻¹

Rule 3: The Constant Multiple Rule

If u is a differentiable function of x and c is a constant, then $\frac{d}{dx}(cu)=c\frac{du}{dx}$

Rule 4: The Sum and Difference Rule

If u and v are differentiable functions of x, then their sum and differences are differentiable at every point where u and v are both differentiable. At such points,

1. $\frac{d}{dx}(u+v)=\frac{du}{dx}+\frac{dv}{dx}$

2. $\frac{d}{dx}(u-v)=\frac{du}{dx}-\frac{dv}{dx}$

Rule 5: The Product Rule

If the products of two differentiable functions u and v is differentiable and...

u = first factor

v = second factor

$\frac{d}{dx}(u\cdot v)=u\cdot \frac{dv}{dx}+v\cdot \frac{du}{dx}$

$\frac{d}{dx}(u\cdot v)=$ (first) · (second)' + (second) · (first)'

Finding Horizontal Tangents:

Horizontal tangents occur when the slope of the derivative equals zero.

Rule 6: The Quotient Rule

At a point where v ≠ 0, the quotient $\frac{u}{v}$ of two differentiable functions is differentiable and...

u = numerator part of quotient
v = denominator part of quotient

$\frac{d}{dx}(\frac{u}{v})=\frac{v\cdot \frac{du}{dx}-u\cdot \frac{dv}{dx}}{v^2}$

Posted: **November 12th, 2011, 12:39 pm**

Derivatives of the Six Trig Functions

$\frac{d}{dx}\sin x=\cos x$

$\frac{d}{dx}\cos x=-\sin x$

$\frac{d}{dx}\tan x=\sec^2 x$

$\frac{d}{dx}\sec x=\sec x\tan x$

$\frac{d}{dx}\cot x=-\csc^2 x$

$\frac{d}{dx}\csc x=-\csc x\cot x$

Posted: **November 16th, 2011, 12:26 am**

Chain Rule

useful when asked to find the derivative of compositions such as $y=\sin(2x-4)$

Method 1: "outside-inside" rule

$y'=\cos(2x-4)$ • ( 2 )
deriv of outside deriv of inside
leave inside alone

$y=2\cos(2x-4)$

Method 2: $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Examples coming soon

Posted: **November 19th, 2011, 10:09 pm**

Finding $\frac{dy}{dx}$ parametrically:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Posted: **December 5th, 2011, 5:05 pm**

Derivatives of Inverse Trig Functions

$\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}u)=\frac{1}{\sqrt{1-u^2}}\bullet \frac{\mathrm{d} u}{\mathrm{d} x}$

$\cos^{-1}u$ is exactly the same except it's negative.

$\frac{\mathrm{d} }{\mathrm{d} x}(\tan^{-1}u)=\frac{1}{1+u^2}\bullet \frac{\mathrm{d} u}{\mathrm{d} x}$

$\cot^{-1}u$ is exactly the same except it's negative.

$\frac{\mathrm{d} }{\mathrm{d} x}(\sec^{-1}u)=\frac{1}{|u|\sqrt{u^2-1}}\bullet \frac{\mathrm{d} u}{\mathrm{d} x}$

$\csc^{-1}u$ is exactly the same except it's negative.

Posted: **December 8th, 2011, 8:42 am**

note: each will be multiplied by $\frac{\mathrm{d} u}{\mathrm{d} x}$

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