Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 425: 4


$y' = \dfrac{1}{4\sqrt{x}+2x}$

Work Step by Step

In order to derivate this function you have to apply the chain rule Let's make an «u» substitution to make it easier $f(u) = \ln(u)$ $u = 2+\sqrt{x}$ Derivate the function: $f'(u) = \dfrac{1}{u} \times u'$ Now let's find u' $u' = \dfrac{1}{2\sqrt{x}}$ Then undo the substitution: $f'(x) = \dfrac{1}{2+\sqrt{x}} \times \dfrac{1}{2\sqrt{x}}$ Simplify and get the answer: $y' = \dfrac{1}{4\sqrt{x}+2x}$
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