# Chapter 13 - The Trigonometric Functions - Chapter Review - Review Exercises - Page 701: 57

$$\frac{{dy}}{{dx}} = - 2x\sin \left( {1 + {x^2}} \right)$$

#### Work Step by Step

\eqalign{ & y = \cos \left( {1 + {x^2}} \right) \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\cos \left( {1 + {x^2}} \right)} \right] \cr & {\text{using the chain rule for }}{D_x}\left( {\tan u} \right) = {\sec ^2}u \cdot {D_x}\left( u \right).{\text{ consider }}u = 9x + 1 \cr & \frac{{dy}}{{dx}} = - \sin \left( {1 + {x^2}} \right)\frac{d}{{dx}}\left[ {1 + {x^2}} \right] \cr & {\text{then}} \cr & \frac{{dy}}{{dx}} = - \sin \left( {1 + {x^2}} \right)\left( {2x} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = - 2x\sin \left( {1 + {x^2}} \right) \cr}

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