Calculus, 10th Edition (Anton)

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

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 56

Answer

$$\frac{d\lambda}{du}=\frac{6(au+b)^5(ad-bc)}{(cu+d)^7}$$

Work Step by Step

$$\frac{d\lambda}{du}=\frac{d}{du}\left(\left(\frac{au+b}{cu+d}\right)^6\right)=6\left(\frac{au+b}{cu+d}\right)^5\frac{d}{du}\left(\frac{au+b}{cu+d}\right)= 6\left(\frac{au+b}{cu+d}\right)^5\frac{\frac{d}{du}(au+b)(cu+d)-(au+b)\frac{d}{du}(cu+d)}{(cu+d)^2}= 6\left(\frac{au+b}{cu+d}\right)^5\frac{a(cu+d)-(au+b)\cdot c}{(cu+d)^2}= 6\left(\frac{au+b}{cu+d}\right)^5\frac{acu+ad-acu-bc}{(cu+d)^2}= 6\left(\frac{au+b}{cu+d}\right)^5\frac{ad-bc}{(cu+d)^2}=\frac{6(au+b)^5(ad-bc)}{(cu+d)^7}$$
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