2.4 Connections F, F' And Trigap Calculus



2.4 connections f f2.4

SERIES EXPANSION OF FUNCTIONS, MACLAURIN’S SERIES, TAYLOR’S SERIES, TAYLOR’S FORMULA. Many functions can be represented by polynomials. In this connection let us note a relationship between the coefficients c 0, c 1, c 2.,c n of the polynomial of degree n. 25=2!2!2!2!2=32 26=2!2!2!2!2!2=64 c. They are half as large each time. Divided by 2, or multiplied by ½ is also acceptable. 20=2!1 2 =1,!2'1=1!1 2 =1 2,2'2=1 2!1 2 =1 2,!2'3=1 4!1 2 =1 8,!2'4=1 8!1 2 =1 16 2'n=1 2n'1!1 21 =1 2n'1+1 =1 2n 1-9. 2!4'22=2!2!!!Check:!1 16 '4=1 4 b. 2!1'2!2=2!3Check:!1 2 '1 4 =1 8 c.

Answer

Work Step by Step

f(x)=(2-x-3$x^{3}$)(7+$x^{5})$ According to the product rule, if f(x)=h(x)g(x), then f'(x)=h(x)g'(x)+g(x)h'(x). If we set h(x)=2-x-3$x^{3}$ and g(x)=7+$x^{5}$, then f'(x)=(2-x-3$x^{3}$)$frac{d}{dx}$[7+$x^{5}$]+(7+$x^{5}$)$frac{d}{dx}$[2-x-3$x^{3}$]. $frac{d}{dx}$[2-x-3$x^{3}]$=-9$x^{2}$-1, using the power rule $frac{d}{dx}$[7+$x^{5}]$=5$x^{4}$, using the power rule Therefore, f'(x)=(2-x-3$x^{3}$)(5$x^{4})$+(7+$x^{5}$)(-9$x^{2}$-1) =10$x^{4}$-5$x^{5}$-15$x^{7}$+(-63$x^{2}$-7-9$x^{7}$-$x^{5})$ =-24$x^{7}$-6$x^{5}$+10$x^{4}$-63$x^{2}$-7

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