How to find f' for f(x) = sin x/ cos x?

Question

Would it be tan x, 1/cos^2 x, or -cos x/sin x?

Answer 1

f(x) = sin(x)/cos(x) is the same as f(x) = tan(x) by the Quotient Identity. It is known that the derivative of tan(x) is sec²(x), so this would be your answer.

You can, however, use quotient rule with sin(x)/cos(x):

d/dx f(x)/g(x) = [g(x)f'(x)-f(x)g'(x)]/(g(x))²

d/dx sin(x)/cos(x) = [cos(x)sin(x)'-sin(x)cos(x)']/(cos(x))² = [cos²(x)+sin²(x)]/(cos²(x)) = 1/cos²(x) = sec²(x)

Note that cos²(x)+sin²(x)=1 by the Pythagorean Identity and 1/cos²(x) = sec²(x) by the Reciprocal Identity.

Hope this helped!

Answer 2

Use the Quotient Rule