How do you derive y=tanx using the definition of the derivative?

Question

Cameron Alexander · Accepted Answer

Use the tangent of a sum, continuity of tangent and #lim_(hrarr0)tan h/h = 1# #f(x) = tanx# #f'(x) = lim_(hrarr0)(tan(x+h) - tanx)/h# # = lim_(hrarr0)((tanx+tan h)/(1-tanxtan h) - tanx)/h# # = lim_(hrarr0)(tanx+tan h- tanx+tan^2xtan h)/(h(1-tanxtan h)) # # = lim_(hrarr0)(tan h/h * (1+tan^2x)/(1-tanxtan h)) # # = (1) * (1+tan^2x)/(1-0) = 1+tan^2x = sec^2x#

Emily Ammerman · Answer

undefined To derive $ y = \tan(x) $ using the definition of the derivative, we start with the definition of the tangent function: $ \tan(x) = \frac{\sin(x)}{\cos(x)} $. Then, we differentiate both sides using the quotient rule for derivatives. The quotient rule states that if $ y = \frac{u}{v} $, then $ y' = \frac{u'v - uv'}{v^2} $, where $ u' $ and $ v' $ denote the derivatives of $ u $ and $ v $ with respect to $ x $ respectively.

Applying the quotient rule to $ \tan(x) = \frac{\sin(x)}{\cos(x)} $, we get:

$$ \frac{d}{dx}\left(\tan(x)\right) = \frac{d}{dx}\left(\frac{\sin(x)}{\cos(x)}\right) $$

$$ = \frac{\cos(x) \cdot \frac{d}{dx}(\sin(x)) - \sin(x) \cdot \frac{d}{dx}(\cos(x))}{(\cos(x))^2} $$

$$ = \frac{\cos(x) \cdot \cos(x) - \sin(x) \cdot (-\sin(x))}{(\cos(x))^2} $$

$$ = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} $$

$$ = \frac{1}{\cos^2(x)} $$

$$ = \sec^2(x) $$

Therefore, the derivative of $ y = \tan(x) $ with respect to $ x $ is $ \sec^2(x) $.