Given that z=sin(x)/cos(x), show that dz/dx = sec^2(x).

We have a "fraction" which we wish to differentiate, so we use the quotient rule with u=sin(x) and v=cos(x).

This means that d/dx of u/v = (vdu/dx - udv/dx)/(v^2).

We have u=sin(x) so du/dx= cos(x).

We have v=cos(x) so dv/dx=-sin(x).

Substitutiong these into the quotient rule formula, we get:

dz/dx = (cos(x)*cos(x) - sin(x)(-sin(x)))/(cos^2(x)).

Double negative in the numerator gives a positive, i.e.

(cos(x)*cos(x) + sin(x)sin(x))/(cos^2(x)).

(cos^2(x) + sin^2(x))/(cos^2(x)).

Now we use the trig identity cos^2(x) + sin^2(x) = 1 to get

dz/dx = 1/cos^2(x)

which we know is the same as sec^2(x) since 1/cos(x) = sec(x).

Therefore dz/dx = sec^2(x).