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by: Hannah Moore

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# Derivatives 201501

Hannah Moore
Clayton State

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Covers derviaties
COURSE
Calculus 1
PROF.
Giovannitti
TYPE
Class Notes
PAGES
3
WORDS
CONCEPTS
chain rule, Product Rule, expondents
KARMA
25 ?

## Popular in Math

This 3 page Class Notes was uploaded by Hannah Moore on Friday March 4, 2016. The Class Notes belongs to 201501 at Clayton State University taught by Giovannitti in Spring 2016. Since its upload, it has received 19 views. For similar materials see Calculus 1 in Math at Clayton State University.

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Date Created: 03/04/16
Derivatives Theroem 2.8 For any real numbers k, m, and b d d d dx(k)=0 dx (m =0 dx(b)=0 Theroem 2.9 Power Rule d (xk=k x k−1 For any nonzero rational number k, dx If f is a linear function that has a slope m and therefor its derivative is constantly m, since the constant identity functions are linear functions with the slope of 0 and 1 their derivatives are 0 and 1 Theorem 2.8 For any real numbers k, m, and b d d d (k)=0 (m =0 (b)=0 dx dx dx k To take the derivative x we bring the exponent k in front of the expression in decreased the exponent by 1 to get KX k-1 Theroem 2.9 Power Rule d k k−1 For any nonzero rational number k, dx(x =k x We can find higher derivatives very easily with this function, for example 4 3 2 (4) (5) f(x)= x f’(x)=4x f’’(x)= 12x f’’’(x)= 24x f (x)= 24 f (x)=0 We know that the limit of a sum is the sum of the limit provided the limits involved exist. The same thing is true for different for derivatives. Theorem 2.10 Derivatives of Constant Multiples and Sums of Functions If f and g are functions k is a constant, then for all x where the functions involued are differentiable, we have the following differentiation formulas: Constant Multiple Rule: (kf)’(x)=kf’x Sun Rule: (f+g)’(x)=f’(x)+g’(x) Difference Rule: (f-g)’(x)=f’(x)-g’(x) While derivatives interact nicely with constant multiples and sums, the same is not true for products and quotients. Theorem 2.11 The Derivatives of Products and Quotients of Function If f and g are functions, then for all x such that both f and g are differentiable, we have the following differentiation formulas: Product Rule: (fg)’(x)= f’(x)g(x)+f(x)g’(x) ' ' ' f x = f (x)g(x)− f x )g(x ) Quotient Rule: (g ( ( )) The derivative of a composition is the product if the derivatives of the component functions. Theorem 2.12 The Chain Rule Suppose f(u(x)) is a composition of functions. For all values of x at which u is differentiable at x and f is differentiable at u(x), the derivative of f with respect to u and the derivative of u with respect to x. df df du = dx du dx ' ' ' (f0g )(x= f u( (x)) (x) 2 2 Although we cannot solve the equation x + y for y and obtain a simple well defined function, we can still think of the x-values as inputs and the y-values as outputs; the only difference is that there may be more than one y-value for each x-value. In cases such as these we say that y is an implicit function of x. Given an equation that defines is an implicit function we can still find the information about slopes and derivatives simply by differentiating both sides of the equation with respect to x. This technique is known as implicit differentiation the key to applying this technique will be applying the chain rule appropriately. So it works like this: x2 +y2=1 <- given equation d/dx (x2 +(y(x))2 ) = d/dx (1) <- differentiate both sides 2x+2y(dy/dx) = 0 <- power and chain rules dy/dx = -2x/2y =-x/y <- Solve for dy/dx Power rule tells us the derivative of x^k is Kx^K -1. This rule does not work in functions where the variable is in the exponent. Theorem 2 13 Derivatives of Exponential Functions For any constant K any constant b>0 with b not = 1 and all x∈ R, d x x a) dx(e =e b x ln¿b b) d x dx(b )=¿ d c) (ek)=ke kx dx

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