User:Bubblechick90210/new article name here

$$\begin{align} & II.Derivatives \\ & A.Definition\text{ }of\text{ }Derivative: \\ & The\text{ derivative of the function }f\text{ with respect to the variable }x\text{ is the function }f'\text{ who value at }x\text{ is:} \\ & f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x+h)-f(x)}{h}\text{                  } \\ & \text{provided the limit exists}\text{.} \\ & \text{The derivative of the function f at the point x=a is the limit} \\ & f'(a)=\underset{x\to a}{\mathop{\lim }}\,\frac{f(x)-f(a)}{x-a} \\ & \text{The domain of f }\!\!'\!\!\text{ may be smaller than the domain of f}\text{.} \\ & \text{When f }\!\!'\!\!\text{ (x) exists, f has a derivative (is differentiable) at x}\text{.} \\ & \text{When a function is differentiable at every point in its domain it is a differentiable function}\text{.} \\ & \text{Ex 1: Find the derivatie of }f(x)=3x^{\text{3}}+4x^{\text{2}}+7\text{ using the defintion of derivative}\text{.} \\ & f'(x)=\underset{h\to 0}{\mathop{\lim }}\,3(x+h)^{3}+4(x+h)^{2}+7-(3x^{3}+4x^{2}+7)/h\text{            Sub (x+h) and (x) where appropriate} \\ & \text{          }\underset{h\to 0}{\mathop{\lim }}\,\text{ }3x^{3}+9x^{2}h+9xh^{2}+3h^{3}+4x^{2}+8xh+4h^{2}+7-3x^{3}-4x^{2}-7/h\text{     Expand} \\ & \text{          }\underset{h\to 0}{\mathop{\lim }}\,9x^{2}h+9xh^{2}+3h^{3}+8xh+4h^{2}/h\text{            Simplify and cancel out like terms}\text{.} \\ & \text{          }\underset{h\to 0}{\mathop{\lim }}\,h(9x^{2}+9xh+3h^{2}+8x+4h)/h\text{              Factor out h} \\ & \text{          }\underset{h\to 0}{\mathop{\lim }}\,9x^{2}+9x(0)+3(0)^{2}+8x+4(0)\text{               Sub in 0 for h} \\ & \text{          }\underset{h\to 0}{\mathop{\lim }}\,9x^{2}+8x \\ & \text{         }f'(x)=9x^{2}+8x \\ & \text{Ex 2: Find the derivative of f(x)=}\sqrt{x+3}\text{ using the definition of derivative at a point}\text{.} \\ & f'(x)=\underset{x\to a}{\mathop{\lim }}\,\sqrt{x+3}-\sqrt{a+3}/(x-a) \\ & \text{           }\underset{x\to a}{\mathop{\lim }}\,\left( \sqrt{x+3}-\sqrt{a+3} \right)\times \left( \sqrt{x+3}+\sqrt{a+3} \right)/(x-a)\times \left( \sqrt{x+3}+\sqrt{a+3} \right) \\ & \text{           }\underset{x\to a}{\mathop{\lim }}\,x+3-(a+3)/(x-a)\times \left( \sqrt{x+3}+\sqrt{a+3} \right) \\ & \text{           }\underset{x\to a}{\mathop{\lim }}\,x-a/(x-a)\times \left( \sqrt{x+3}+\sqrt{a+3} \right) \\ & \text{           }\underset{x\to a}{\mathop{\lim }}\,1/\left( \sqrt{a+3}+\sqrt{a+3} \right) \\ & \text{           }\underset{x\to a}{\mathop{\lim }}\,1/\left( 2\sqrt{a+3} \right) \\ & \text{B}\text{. Interpretations of the Derivative and Differentiability} \\ & \text{There are right-hand and left hand derivatives of a function, denotated as + from the right or - from the left}\text{. The right and left-hand limits must exist and be the same at a point for the function to be differentiable at that point}\text{.} \\ & \text{If f has a derivative a x=a, then f is continous at x=a} \\ & \text{A function might fail to be differentiable at a point where there is a corner, a cusp, a vertical tangent, or a discontinuity}\text{.} \\ & \text{Ex: Find all points in the domain of f(x)=}\left| x-3 \right|+7\text{ where f is not differentiable}\text{.} \\ & \text{The graph of this function is the same as that of y=}\left| x \right|,\text{ translated 3 units to the right and 7 units up}\text{. This puts the corner at the point (3,7), so this function is not differentiable at x=3}\text{. At every other point the graph is differentiable}\text{.} \\ & \text{C}\text{. Average Rates of Change Vs}\text{. Instantaneous Rates of Change} \\ & \text{The average rate of change of a function is defined as }f(t+\Delta t)-f(t)/\Delta t.\text{ It is the change in y over the change in x, or in other words it is the slope of the secant to the curve through the points (t, f(t)) and (t+}\Delta t,f(\text{t+}\Delta t).\text{ } \\ & \text{The instantaneous rate of change is defined as }\underset{h\to 0}{\mathop{\lim }}\,f(x+h)-f(a)/h,\text{or in other words, we take the derivative of the average rate of change}\text{. This is the slope of the curve at (t, f(t))}\text{.} \\  & \text{Ex:Let y=x}^{2}-2 \\  & \text{(a) Find the average rate of change of  with respect to over the interval}\left[ 2,5 \right] \\  & \text{(b) Find the instantaneous rate of change of with respect to x at the point x=4}\text{.} \\  & \text{Solution:} \\  & \text{(a) For Average Rate of Change:  f(5)=5}^{2}-2=23\text{       f(2)=2}^{2}-2=2 \\  & \frac{23-2}{5-2}=\frac{21}{3}=7 \\ & \text{(b) For Instantaneous Rate of Change: } \\ & \underset{h\to 0}{\mathop{\lim }}\,(x+h)^{2}-2-(\text{x}^{2}-2)/h \\ & \underset{h\to 0}{\mathop{\lim }}\,x^{2}+2xh+h^{2}-2-\text{x}^{2}+2/h \\ & \underset{h\to 0}{\mathop{\lim }}\,2xh+h^{2}/h \\ & \underset{h\to 0}{\mathop{\lim }}\,h(2x+h)/h \\ & \underset{h\to 0}{\mathop{\lim }}\,2x+h \\ & f'(x)=2x \\ & 2(4)+0=8\text{                   Sub in 4 for x}\text{.} \\ & \text{D}\text{. Equation of Tangent Lines and Normal Lines} \\ & \text{The tangent line of a function occurs where the slope, }f'(x),\text{ is equal to zero}\text{. Once you know the equation of the tangent line, you can find the equation of the normal line by take the opposite reciprocal of the slope because the normal line is perpendicular to } \\ & \text{the tangent line}\text{.} \\ & \text{Ex: Find the eqautions of the tangent and normal lines of f(x) =}x^{3}+7x^{2}+2x+11\text{ at (2,51)} \\ & f'(x)=3x^{2}+14x+2 \\ & f'(2)=42 \\ & y-y_{1}=m(x-x_{1}) \\ & \text{Tangent Line                   Normal Line} \\ & y-51=42(x-2)\text{            }y-51=\frac{-1}{42}(x-2) \\ \end{align}$$

$$\begin{align} & \text{E}\text{. Derivative Formulas} \\ & \text{Constant Functions: }\frac{d}{dx}(c)=0\text{    Ex:}\frac{d}{dx}e=0 \\ & \text{Power Functions:}\frac{d}{dx}(x^{n})=nx^{n-1}\text{ Ex:}\frac{d}{dx}5x^{4}-2x^{3}+6x^{2}-8x+2=20x^{3}-6x^{2}+12x \\ & \text{Exponential Functions:}\frac{d}{dx}(e^{x})=e^{x}\text{    }\frac{d}{dx}(a^{x})=a^{x}\ln a\text{     Ex:}\frac{d}{dx}5^{2x+3}=5^{2x+3}\times \ln 5 \\ & \text{Logarithmic Functions:}\frac{d}{dx}(\ln x)=\frac{1}{x}\text{  Ex: }\frac{d}{dx}(\ln e)=\frac{1}{\ln e}=1 \\ & \text{Trigonometric Functions:}\frac{d}{dx}(\sin x)=\cos x\text{      }\frac{d}{dx}(\cos x)=-\sin x\text{       }\frac{d}{dx}(\tan x)=\sec ^{2}x\text{      }\frac{d}{dx}\csc x=-\csc x\cot x\text{      }\frac{d}{dx}\sec x=\sec x\tan x\text{       }\frac{d}{dx}\cot x=-\csc ^{2}x\text{        Ex:}\frac{d}{dx}\cot (x-5)=-\csc ^{2}(x-5) \\ & \\  & \text{Inverse Trigonometric Functions: }\frac{d}{dx}(\sin ^{-1}x)=\frac{1}{\sqrt{1-x^{2}}}\text{         }\frac{d}{dx}(\cos ^{-1}x)=-\frac{1}{\sqrt{1-x^{2}}}\text{       }\frac{d}{dx}(\tan ^{-1}x)=\frac{1}{1+x^{2}}\text{        }\frac{d}{dx}(\csc ^{-1}x)=-\frac{1}{x\sqrt{x^{2}}-1}\text{      }\frac{d}{dx}(\sec ^{-1}x)=\frac{1}{x\sqrt{x^{2}}-1}\text{    }\frac{d}{dx}(\cot ^{-1}x)=-\frac{1}{1+x^{2}}\text{  }\frac{d}{dx}\cot ^{-1}(x-5)=-\frac{1}{x^{2}-10x+26} \\ & \text{F}\text{. Derivatives of Inverse Functions:} \\ & \text{If }f\text{ is differentiable at every point of an interval }I\text{ and }df/dx\text{ is never zero on }I\text{, then }f\text{ has an inverse and }f^{-1}\text{ is differentiable at every point of the interval }f(I)\text{.} \\ & \text{Ex: Let f(x)}=\cos \text{ }x\text{ }+3x \\ & \text{a) Find the derivative of }f(x):\text{     }f'(x)=-\sin \text{ }x+3 \\  & \text{b) Find }f(0)\text{ and }f'(0):\text{     }f(0)=1\text{    }f'(0)=3 \\ & \text{c) Find }f^{-1}(1)\text{ and (}f^{-1})'(1).\text{   }f^{-1}(1)=0\text{   (}f^{-1})'(1)=\frac{1}{3} \\  & \text{G}:\text{Rules for Differentiation} \\  & \text{Sum/Difference Rule: }\frac{d}{dx}(u\pm v)=\frac{du}{dx}\pm \frac{dv}{dx}\text{    } \\  & \text{Ex: }\frac{d}{dx}\text{3x}^{2}+\sin x+7x+3=6x+\cos x+7 \\  & \text{Constant Multiplier Rule:} \\  & \frac{d}{dx}(cu)=c\frac{du}{dx}\text{      } \\  & \text{Ex:}\frac{d}{dx}7\tan x=7\sec ^{2}x \\  & \text{Product Rule:} \\  & \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx} \\  & \text{Ex: }\frac{d}{dx}(x{}^{2}+7)(x^{3}+2x) \\  & \text{      (}x{}^{2}+7)(3x^{2}+2)+(x^{3}+2x)(2x)=5x^{4}+27x^{2}+14 \\  & \text{Quotient Rule: }\frac{d}{dx}\left( \frac{u}{v} \right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}} \\  & \text{Ex:}\frac{d}{dx}\frac{x{}^{2}+7}{x^{3}+2x}=\frac{(x^{3}+2x)(2x)-\text{(}x{}^{2}+7)(3x^{2}+2)}{(x^{3}+2x)^{2}}=\frac{2x^{4}+4x^{2}-3x^{4}-2x^{2}-21x^{2}-14}{(x^{3}+2x)^{2}}=\frac{-x^{4}-19x^{2}-14}{(x^{3}+2x)^{2}} \\ & \text{Chain Rule: }(f\circ g)'(x)=f'(g(x))\text{ }\centerdot \text{ }g'(x) \\ & \text{Ex:}\frac{d}{dx}\ln \left( \sin ^{2}(\frac{1}{x}) \right)=\frac{1}{\sin ^{2}(\frac{1}{x})}\centerdot 2\sin (\frac{1}{x})\centerdot \cos (\frac{1}{x})\centerdot \frac{-1}{\sqrt{x}}=\frac{-2\cos (\frac{1}{x})}{\sqrt{x}\sin \frac{1}{x}} \\ & \text{H}\text{. Higher-Order Derivatives and Derivative Notations} \\ & \text{There are many ways to denote the derivatice of a function y=f(x)}\text{. Some include:} \\ & \text{f }\!\!'\!\!\text{ (x)} \\ & \text{y }\!\!'\!\!\text{ } \\ & \frac{dy}{dx} \\ & \frac{df}{dx} \\ & \frac{d}{dx}f(x) \\ & \text{Higher-order derivatives can be the second derivative, third derivative, fourth derivative, and so on}\text{. The second derivative can be notated as f }\!\!'\!\!\text{ }\!\!'\!\!\text{ (x) or as }\frac{\text{d}^{2}y}{dx^{2}}\text{ and other higher-order derivatives in the same manner}\text{.} \\ & \text{Knowing the second derivative can help you determine things such as the concavity of a function}\text{.} \\ & \text{Ex: Find the first, second, and third derivative of f(x)=7}x^{4}+3x^{3}+\sin x \\ & f'(x)=28x^{3}+9x^{2}+\cos x \\ & f''(x)=84x^{2}+18x-\sin x \\ & f'''(x)=168x+18-\cos x \\ & \text{I}\text{. Implicit Differentiation:} \\ & \text{Implicit differentiation is used to find the derivative when y is not defined explicitly in terms of x but is diffrentiable}\text{. To solve for }\frac{dy}{dx},\text{differentiate both sides with respect to x}\text{.} \\ & \text{Ex: Find }\frac{dy}{dx}\text{ if x}^{2}-2xy+3y^{2}=2 \\ & \text{                       2x-2}\left( x\frac{dy}{dx}+y\centerdot 1 \right)+6y\frac{dy}{dx}=0 \\ & \text{                      2x-2x}\frac{dy}{dx}-2y+6y\frac{dy}{dx}=0 \\ & \text{                      }\frac{dy}{dx}\left( -2x+6y \right)=2y-2x \\ & \text{                       }\frac{dy}{dx}=\frac{y-x}{3y-x} \\ & \text{J}\text{.Logarithmic Differentiation:} \\ & \text{Ex:Differentiate y=}\frac{x^{\tfrac{4}{3}}\centerdot \sqrt{x^{2}+1}}{(4x+5)^{6}} \\ & \ln y=\ln x^{\tfrac{4}{3}}+\ln \sqrt{x^{2}+1}-\ln (4x+5)^{6}\text{            Take the log of both sides} \\ & \ln y=\frac{4}{3}\ln x+\frac{1}{2}(x^{2}+1)-6\ln (4x+5)\text{          Property of logs} \\ & \frac{1}{y}\frac{dy}{dx}=\frac{4}{3x}+\frac{x}{x^{2}+1}-\frac{24}{4x+5}\text{                         Differentiate implicitly} \\ & \frac{dy}{dx}=y\left( \frac{4}{3x}+\frac{x}{x^{2}+1}-\frac{24}{4x+5}\text{ } \right)\text{                    Multiple both side by y}\text{.} \\ & \frac{dy}{dx}=\left( \frac{x^{\tfrac{4}{3}}\centerdot \sqrt{x^{2}+1}}{(4x+5)^{6}} \right)\left( \frac{4}{3x}+\frac{x}{x^{2}+1}-\frac{24}{4x+5}\text{ } \right)\text{ Sub the original equation back in for y}\text{.} \\ & \\ \end{align}$$