Answer
$p^{\prime}=sec^2q$
Work Step by Step
As given that
$p=5+\frac{1}{cotq}$
$p=5+tanq$ (because tan $\theta$ and $cot \theta$ are reciprocals)
We differentiate both sides to obtain:
$p^{\prime}=0+sec^2q$
$p^{\prime}=sec^2q$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 27
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