Derivative of tanx?

1 Answer
May 2, 2018

#\frac{1}{cos^2(x)}#

Explanation:

You may also consider #tan(x)# as a generic function defined by ratio:

#f(x) = \frac{f(x)}{g(x)}#

and apply the rule

#f'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}#

In your case,

#tan(x) = \frac{sin(x)}{cos(x)}#

and thus the rule states that

#\frac{d}{dx}tan(x) = \frac{d}{dx}\frac{sin(x)}{cos(x)} = \frac{cos(x)cos(x) - sin(x)(-sin(x))}{cos^2(x)}#

which evaluates to

#\frac{cos^2(x)+sin^2(x)}{cos^2(x)} = \frac{1}{cos^2(x)}#