How do you differentiate #f(x)=tanx# using the quotient rule?

Question

Samantha Adkison · Accepted Answer

#f'(tanx)=sec^2x#

Using the trig identities, recall that #tanx# can be rewritten as #sinx/cosx#. Now differentiate using quotient rule:

#d/dx(f(x)/g(x)) = (g(x)*f'(x) - f(x)*g'(x))/g(x)^2#

#f'(x)=(cosx*cosx-sinx*-sinx)/(cos^2x)#

#f'(x)=(cos^2x+sin^2x)/cos^2x#

Recall from the trig identities that #cos^2x+sin^2x=1#. Simplify:

#f'(x)=1/cos^2x#

Finally, simplify again with trig identities and the result is:

#f'(x)=sec^2x#

Luke Alvarez · Answer

undefined To differentiate $ f(x) = \tan(x) $ using the quotient rule, we first express tangent as the quotient of sine and cosine: $ \tan(x) = \frac{\sin(x)}{\cos(x)} $. Then, applying the quotient rule $ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} $, where $ u = \sin(x) $ and $ v = \cos(x) $, we have:

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

Differentiating sine and cosine yields $ \frac{d}{dx}(\sin(x)) = \cos(x) $ and $ \frac{d}{dx}(\cos(x)) = -\sin(x) $, so:

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

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

Using the Pythagorean identity $ \cos^2(x) + \sin^2(x) = 1 $, we simplify to:

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

Which can be rewritten as:

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