## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 3 - Derivatives - Review Exercises - Page 234: 54

#### Answer

$$1$$

#### Work Step by Step

\eqalign{ & \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^2}\left( {\frac{\pi }{4} + h} \right) - \frac{1}{2}}}{h} \cr & {\text{rewrite}} \cr & \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^2}\left( {\frac{\pi }{4} + h} \right) - \frac{1}{2}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^2}\left( {\frac{\pi }{4} + h} \right) - {{\sin }^2}\left( {\frac{\pi }{4}} \right)}}{h} \cr & {\text{using the definition of the derivative }}f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} \cr & {\text{comparing }}\mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^2}\left( {\frac{\pi }{4} + h} \right) - {{\sin }^2}\left( {\frac{\pi }{4}} \right)}}{h}{\text{ with the definition of the derivative of a point }}a \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^2}\left( {\frac{\pi }{4} + h} \right) - {{\sin }^2}\left( {\frac{\pi }{4}} \right)}}{h}:\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} \cr & {\text{we can note that }}f\left( x \right) = {\sin ^2}x{\text{ and }}a = \frac{\pi }{4} \cr & {\text{then}} \cr & \,\,\,\,\,\,\,\,f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\sin }^2}x} \right] \cr & \,\,\,\,\,\,\,\,f'\left( x \right) = 2\sin x\cos x \cr & \,\,\,\,\,\,\,\,f'\left( {\frac{\pi }{4}} \right) = 2\sin \left( {\frac{\pi }{4}} \right)\cos \left( {\frac{\pi }{4}} \right) \cr & \,\,\,\,\,\,\,\,f'\left( {\frac{\pi }{4}} \right) = 1 \cr & {\text{therefore}}{\text{,}} \cr & \,\,\,\,\,\,\,\mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^2}\left( {\frac{\pi }{4} + h} \right) - \frac{1}{2}}}{h} = f'\left( {\frac{\pi }{4}} \right) = 1 \cr}

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