Derivative+Rules+for+All+Functions

__Derivative Rules for All Functions__ **If possible, try to include some images or other materials to enhance the wiki.**

Power, Trig, logarithmic, exponential, inverse trig, inverse functions

**__Power Rule__**: This rule is by far the simplest of the deriving shortcuts. First we start with the equation: x^3+4x^2+20x+45 In order to derive this, take the power that the x is exponentiated by, bring it down and multiply it by that number, then decrease the exponent by one. You do this for each term and you get the new equation: 3x^2+8x+20 If the x's exponent is one, just remove the x, and if there are stray numbers without an x, that goes to zero. The Power rule can be modeled like this: f(x)= x^n y' = nx^(n-1) This also works with the chain rule take the equation y = 8(x^2+3)^3 y' = (8)(3)(2x)(x^2+3)^2 Where you multiple by the inside's derivative as well as following the basic power rule formula.

__ **Trigonometric Function Rules**: __ Sine and cosine functions have derivatives that relate to each other which makes them easy to memorize. If y(x) = sin(x) then y’(x) = cos(x) If y(x) = cos(x) then y’(x) = -sin(x) If y(x) = -sin(x) then y’(x) = -cos(x) If y(x) = -cos(x) then y’(x) = sin(x) Notice a pattern? The other trig functions have derivatives that are found using trig identities and the quotient rule. If y(x) = tan(x) then y’(x) = sec(x)^2 because tan(x) = sin(x)/cos(x). With the quotient rule we see that y’(x) = [sin(x)^2+cos(x)^2]/cos(x)^2 = 1/(cos(x)^2) = sec(x)^2. If y(x) = csc(x) then y’(x) = -csc(x)cot(x) because csc(x) = 1/sin(x). With the quotient rule we see that y’(x) = -cos(x)/[sin(x)^2] = [-1/sin(x)]*[cos(x)/sin(x)] = -csc(x)cot(x). If y(x) = sec(x) then y’(x) = sec(x)tan(x) because sec(x) = 1/cos(x). With the quotient rule we see that y’(x) = sin(x)/[cos(x)^2] = [1/cos(x)]*[sin(x)/cos(x)] = sec(x)tan(x). If y(x) = cot(x) then y’(x) = -csc(x)^2 because cot(x) = cos(x)/sin(x). With the quotient rule we see that y’(x) = [-sin(x)^2-cos(x)^2]/sin(x)^2 = -1*(1/sin(x)^2) = -csc(x)^2. Remember to apply the chain rule for these also when applicable. If y(x) = sec(2x) then y’(x) = 2*sec(2x)tan(2x). __ **Logarithm Function Rules**: __ Log rules are fairly simple. The derivative of log(x) is 1/((x)(ln10). The derivative of ln(x) is 1/x. If y(x) = log(x) then y’(x) = 1/((x)(ln10) Any coefficient in front of log will be multiplied by the derivative of the log to create the derivative of the function. If y(x) = 9*log(x) then y’(x) = (9)*1/((x)(ln10) = 9/((x)(ln10)) If there is an expression other than x inside the log, the chain rule must be applied. If y(x) = log(5x^2+3) then y’(x) = (10x)*1/((5x^2+3)(ln10)) = (10x)/((5x^2+3)(ln10)) Because of this, any log function with a linear expression inside of it will derive to 1/x also. If y(x) = log(3x) then y’(x) = [1/((3x)(ln10)]*3 = 3/((3x)(ln10)) = 1/((x)(ln10) The model is y = logbA where b is the log base, and A is the number that is "logged". y' = 1/((A)(lnb))*(da/dy)<==(derivative of a, basically including the chain rule.) y = x^a y' = (lnx)(x^a)(da/dx)<=( the derivative of a) For example! y = 10^x y' = (ln10)(10^x) or y = 5^2x y' = (2)(ln5)(5^2x) As you see, the chain rule is omni-present when it comes to deriving. e^x is a type of exponential function. It works the same way, however; the when derived you get e^x because: y = e^x y' = (e^x)(lne) <== equals 1, and the derivative of x is just one as well. but if y = e^(x^2) y' = 2x(e(x^2)) __ **Inverse Trigonometric Functions Rules**: __ These derivatives don’t really make sense if you try to figure them out with the rules described above so just memorize them (they require implicit differentiation). If y(x) = arcsin(x) = then y’(x) = 1/ √(1-x^2). If y(x) = arccos(x) then y’(x) = -1/ √(1-x^2). If y(x) = arctan(x) then y’(x) = 1/ (1+x^2). For these, if the inverse trig function has an inside function you replace x with it in the derivative. Once again, remember to use chain rule when needed. If y(x) = arctan(3x) then y’(x) = 3*1/ [1+(3x)^2] = 3/(1+9x^2). If y(x) = arctan(x^3) then y’(x) = 3x^2*1/[1+(x^3)^2] = 3x^2/ (1+x^6). **__Inverse Function Rules!:__** For the inverse function we will start with the function y = x^2 First swap the y and x values for eachother to get x = y^2. Then Solve for y to get y = x^(1/2) Finally derive both sides to get the inverse function derivative of y' = (1/2)(x)^(-1/2)
 * __Exponential Function Rules:__** for the exponential function the model looks like this:

Lets try it with an exponential function! y = 4^x x = 4^y log 4 x = y y' = 1/((ln4)(x))

Finally lets make this somewhat interesting. y = (3x-2)^4 + 5 x = (3y-2)^4 + 5 x-5 = (3y-2)^4 (x-5)^(1/4) = 3y - 2 (1/3)((x-5)^(1/4) + 2) = y y' = (1/12)(x-5)^(-3/4) There you have it. That is how you derive inverse functions.

Topic: Derivative Rules Group Members: Brandan and Marvin

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The notation used is not acceptable. Don’t use ^2 or * for multiplication. Use an equation editor. No images or media aids to help develop your topic. Could also show derivation of inverse trig derivatives using implicit differentiation. Or show proof of derivatives of exponentials using the definition of derivative. Organization could be better also.

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