Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.2 Exercises: 35


The slope at the point $(0, 1)$ is 8.

Work Step by Step

To workout this problem, you need to first expand the function using F.O.I.L. ( a quick review of expanding: Multiply the first terms together then the outer terms then the inner terms and finally the last terms; therefore in this case we get $(4x+1)^2 = (4x+1)(4x+1) = (4x)(4x)+(4x)(1)+(1)(4x)+(1)(1) = 16x^2+8x+1$ After expanding we get that $y = 16x^2+8x+1$. Now, the problem is just simple application of the power rule $((x^n)'=nx^{n-1})$ on the multiple terms to get the overall derivative. The derivative of $16x^2$ is $(16)(2)(x^{2-1})$ which is $32x$. The derivative of $8x$ is equal to $(8)(1)(x^{1-1})$ which is 8. The derivative of $1$ is $0$. Adding everything together, we get the total derivative to be $32x+8$. To find the slope at the point, plug in the x-coordinate to get $(32)(0)+8$ which is equal to $8$. N.B. In section 2.4, you will take the chain rule which will allow to you calculate derivatives of functions similar to the above raised to powers such as 10 and 11 (you won't be expanding thirteen terms for sure!).
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