Using the limit definition, how do you find the derivative of #f(x) = (sec x) #?

Question

Ezekiel Alexander · Accepted Answer

Until the questions can be merged, here is the answer

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Luke Alvarez · Answer

undefined To find the derivative of $ f(x) = \sec(x) $ using the limit definition, we start with the definition of the derivative:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute $ f(x) = \sec(x) $ into the definition:

$$ f'(x) = \lim_{h \to 0} \frac{\sec(x + h) - \sec(x)}{h} $$

Apply the identity $ \sec(x) = \frac{1}{\cos(x)} $:

$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\cos(x + h)} - \frac{1}{\cos(x)}}{h} $$

To simplify, multiply the numerator and denominator by $ \cos(x + h) \cos(x) $:

$$ f'(x) = \lim_{h \to 0} \frac{\cos(x) - \cos(x + h)}{h \cos(x) \cos(x + h)} $$

Apply the trigonometric identity $ \cos(A) - \cos(B) = -2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right) $:

$$ f'(x) = \lim_{h \to 0} \frac{-2 \sin \left( \frac{x + (x + h)}{2} \right) \sin \left( \frac{x + (x + h)}{2} \right)}{h \cos(x) \cos(x + h)} $$

Simplify further:

$$ f'(x) = \lim_{h \to 0} \frac{-2 \sin(x + \frac{h}{2}) \sin(\frac{h}{2})}{h \cos(x) \cos(x + h)} $$

As $ h $ approaches 0, $ \sin(\frac{h}{2}) $ approaches 0, so we can rewrite the limit as:

$$ f'(x) = \frac{-2 \sin(x) \cdot 0}{0 \cdot \cos(x) \cos(x)} $$

This expression is in an indeterminate form of $ \frac{0}{0} $, so we can apply L'Hôpital's Rule:

$$ f'(x) = \lim_{h \to 0} \frac{\frac{d}{dx} (-2 \sin(x + \frac{h}{2}) \sin(\frac{h}{2}))}{\frac{d}{dx} (h \cos(x) \cos(x + h))} $$

$$ f'(x) = \lim_{h \to 0} \frac{-2 \cos(x + \frac{h}{2}) \sin(\frac{h}{2})}{-\sin(x) \cos(x + h) - h \sin(x) \sin(x + h)} $$

$$ f'(x) = \frac{-2 \cos(x) \sin(0)}{-\sin(x) \cos(x) - 0} $$

$$ f'(x) = \frac{-2 \cos(x) \cdot 0}{-\sin(x) \cos(x)} $$

$$ f'(x) = 0 $$

Therefore, the derivative of $ f(x) = \sec(x) $ is $ f'(x) = 0 $.