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# Chain rule and derivatives of Exponential Functions Math 120-4

EMU

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This 4 page Class Notes was uploaded by Megan steltz on Saturday October 15, 2016. The Class Notes belongs to Math 120-4 at Eastern Michigan University taught by John G. Patrick in Fall 2016. Since its upload, it has received 3 views. For similar materials see Calculus I in Math at Eastern Michigan University.

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Date Created: 10/15/16

Chain Rule We need to find derivatives of compositions of 2 or more functions. Ex. f(x) = sin(x^2), g(x)= x + 1, h(x)= cos(sintan(x)) The chain rule F(x)=f(g(x)), F’(x)=f’(g(x))*g’(x) Ex. F(x) = sin(x^2) the other function f(g(x)) is sin(g(x)) the inner function g(x) is x^2 F’(x) = cos(x^2)*d/dx x^2,cos(x^2)*2x Ex. g(x) = x + 1, d/dx( x + 1) = d/dx (x^2+1)^½ The outer function is the power of ½, the inner function is x^2 +1 d/dx [ x^2 +1)^½] = ½(x^2+1)^-½* d/dx(x^2+1) = ½(X^2+1)^-½*2x this answer is perfectly acceptable Ex. d/dx [(x^4+x)^10] = 10(x^4+x)^9 * 4x^3 +1 Ex. d/dx (sin^8(x)) Remember sin^n(x) = (sin(x))^n So d/dx [(sin(x))^8] =8(sin(x))^7 * cos(x) Mini Formula = d/dx ((f(x)^n) = n/f(x))^n-1 * f’(x) Advice always rewrite trig functions like the equation above Ex. d/dx sec(x^2) = sec(x^2)tan(x^2) * 2x Ex. d/dx [sec(sin^2(x^3))] = d/dx [sec((sin(x^3))^2] Composition of 4 functions [sec((sin(x^3))^2][tan((sin(x^3))^2] * 2(sin(x^3))*cos(x^3)*3x^2 The next function is to the power of two Ex. d/dx sincos(tan(x)) =cos(cos(tan(x)))*-sin(tan(x))*sec^2(x) Ex. d/dx ( 3 1 ) ((sin(x))^5)^-⅓ √ sin (x) = sin(x)^-5/3 = -5/3(sin(x))^-8/3 * cos(x) Exponential Functions Are of the form f(x) + a^x where a>0 , a/ 1 Not to be confused with a power function like x^a In an exponential growth model, the growth rate is proportional to the size of the function. We expect that d/dx (a^x) = K*a^x for some constant To find d/dx(2^x) 2x+h−2 x d/dx (2^x) = lim( h ) h ⇒0 2 2 −2 x = lim( h ) h ⇒0 h = lim(2^x (2 −1) ) h h ⇒0 (2 −1) = lim (2^x) li( ) h h ⇒0 h ⇒0 (2 −1) = 2^x * lim( h ) h ⇒0 I.e. the original function times some constant To finish solving graph y= (2^x -1)/x near x = 0 lim ≈ .69 h ⇒0 h We can see that lim ( (2 −1) ) ≈ .69 h h ⇒0 So d/dx (2^x) ≈.69 * 2^x h Similarly d/dx (3^x) = 3^x * li((3 −1) ) ≈1.1 * 3^x h h ⇒0 So we’ve shown that the derivative of an y exponential function is a constant times the function h Def. e≈ 2.71828… is the unique number for which lim (e −1) )= 1 h h ⇒0 (e −1) So we have d/dx e^x = e^x lim ( h ) = e^x h ⇒0 The growth rate of e^x equals the value of the function d/dx = e^5x = e^5x d/dx (5x) = 5e^5x The exponential (e) is the outer function; the power (5x) is the inner function d/dx (e^kx) = ke^kx d/dx(e^-x) = -e^-x d/dx(e^x^2 +2x) = e^x^2 + 2x (2x +2) Mini rule: d/dx e^f(x) = e^f(x) * f’(x) Recall: e^x and ln(x) are inverse functions ln(e^x) = x e^ln(x) = x ln(e) = 1 ln(e^2)=2 ln(1)=0 To find d/dx (2^x), note that 2 =e^ln(2) d/dx(2^x) = d/dx((e^ln(2))^x) = d/dx(e^ln(2)*x By the chain rule = (e^ln(2)*x)*d/dx (ln(2)*x) = e^ln(2) *ln(2) = ln(2)(e^ln(2))^x = ln(2)*2^x Thm d/dx a^x = ln(a)*a^x In the case where the base is e, d/dx e^x = ln(e)e^x = e^x

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