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Differentiating and Integrating Mathematical Functions: Solutions and Examples, Assignments of Calculus

Solutions and explanations for differentiating and integrating various mathematical functions using the chain rule, product rule, and substitution method. It covers functions such as sin(x), log(x), tan(x), and exponential functions.

Typology: Assignments

Pre 2010

Uploaded on 09/17/2009

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koofers-user-4pj 🇺🇸

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Download Differentiating and Integrating Mathematical Functions: Solutions and Examples and more Assignments Calculus in PDF only on Docsity! Jim LambersMath 2BFall Quarter 2004-05Homework Assignment 5 Solution1. Di erentiate the following fun tions with respe t to x.(a) f(x) = 2sinxSolution Using the rule ddx [ax℄ = ax lna;in onjun tion with the Chain Rule, we obtainf 0(x) = 2sinx ln 2 ddx [sinx℄ = 2sinx ln 2 os x:(b) f(x) = log3(x22x). Di erentiate without using the Produ t Rule.Solution First, we use the properties of logarithmi fun tions,logb(xy) = logb x+ logb y; logb xy = y logb xto obtain f(x) = log3 x2 + log3 2x = 2 log3 x+ x log3 2:We then use the rule ddx [logb x℄ = ddx  lnxln b  = 1x ln bto obtain f 0(x) = 2x ln 3 + log3 2:( ) f(x) = sin1( s x)Solution Using the rulesddx [sin1 x℄ = 1p1 x2 ; ddx [ s x℄ = s x ot x;in onjun tion with the Chain Rule, we obtainf 0(x) = 1p1 s 2 x ddx [ s x℄ = s x ot xp1 s 2 x:1 (d) f(x) = se (tan1 x)Solution Using the rulesddx [se x℄ = se x tan x; ddx [tan1 x℄ = 11 + x2 ;in onjun tion with the Chain Rule, we obtainf 0(x) = se (tan1 x) tan(tan1 x) ddx [tan1 x℄ = se (tan1 x) x1 + x2 :The x in the numerator arises from the an ellation equation tan(tan1 x) = x.2. Evaluate the following de nite integrals. Please simplify your answer as mu h as possible.(a) R 20 x10x2 dxSolution Using the substitution u = x2, for whi h du = 2x dx, we obtainZ 20 x10x2 dx = 12 Z 40 10u du= 12 10uln 10 40= 12 ln 10(104 100)= 99992 ln 10 :(b) R 30 10log100 x2 dx. Hint: use the hange-of-base formula and the fa t that ax is the inverseof loga x. Then the problem will be ome very easy. Really.Solution First, we rewrite the integrand using the hange-of-base formulalogb u = loga uloga b ;with u = x2, b = 100, and a = 10. This yieldslog100 x2 = log10 x2log10 100 = 2 log10 x2 = log10 x:It follows from the an ellation equationblogb x = x2
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