Calculus with Applications (10th Edition)

Published by Pearson

Chapter 13 - The Trigonometric Functions - Chapter Review - Review Exercises - Page 701: 53

Answer

$$\frac{{dy}}{{dx}} = 6x{\csc ^2}\left( {6 - 3{x^2}} \right)$$

Work Step by Step

\eqalign{ & y = \cot \left( {6 - 3{x^2}} \right) \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\cot \left( {6 - 3{x^2}} \right)} \right] \cr & {\text{using the chain rule for }}{D_x}\left( {\cot u} \right) = - {\csc ^2}u \cdot {D_x}\left( u \right).{\text{ for this exercise }}u = 6 - 3{x^2} \cr & {\text{then}} \cr & \frac{{dy}}{{dx}} = - {\csc ^2}\left( {6 - 3{x^2}} \right)\frac{d}{{dx}}\left[ {6 - 3{x^2}} \right] \cr & {\text{solve derivative}} \cr & \frac{{dy}}{{dx}} = - {\csc ^2}\left( {6 - 3{x^2}} \right)\left( { - 6x} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = 6x{\csc ^2}\left( {6 - 3{x^2}} \right) \cr}

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