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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 34e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 34e

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# A derivative formula a. Use the definition of the

ISBN: 9780321570567 2

## Solution for problem 34E Chapter 3.1

Calculus: Early Transcendentals | 1st Edition

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Problem 34E

A derivative formula a?. Use the definition of the derivative to determine d(?ax + b , where a ? ? and ?b dx are constants. d b.? Use the result of part (a) to find dx(?5x + 9 .

Step-by-Step Solution:
Step 1 of 3

SOLUTION STEP 1 (a). Here f(x) = ax + b We have to find df(x) = f(x) dx f(x+h)f(x) The definition of f(x) = lim h h0 Thus on substituting the given values into the definition,we get a(x+h)+b ax+b f(x) = lim h0 h To simplify this,we will multiply the numerator and the denominator with the conjugate of the numerator. Thus we get a(x+h)+b ax+b a(x+h)+b ax+b a(x+h)+b+ ax+b f(x) = lim h = lim h × h0 h0 a(x+h)+b+ ax+b 2 2 Here,using the formula (a + b)(a b) = (a b ) in the numerator, we can write the above equation as (a(x+h)+b)(ax+b) lim = lim ah = lim a h0 h(a(x+h)+b+ ax+b h0 h a(x+h)++ ax+b) h0 a(x+h)+b ax+b) On taking the limit,we get a a f(x) = ax+b+ax+b)= 2 ax+b a Therefore f x) = 2 ax+b STEP 2 (b). Here,using the above result ,we need to find the value of d (5x + 9 . dx We got, dx( ax + b) = a 2ax+b Therefore, d(5x + 9= 5 dx 2 5x+9 a 5 f x) = = 2 ax+b 2 5x+9

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