In the chapter on trigonometric functions we have seen that #\tan(x)# is defined as #\frac{\sin(x)}{\cos(x)}#. Now that we know the derivatives of #\sin(x)# and #\cos(x)#, we can determine the derivative of #\tan(x)# using the quotient rule, which we will talk about later.

An alternative proof makes use of the product rule and the chain rule. We write \[f(x)=\tan(x) = \dfrac{\sin(x)}{\cos(x)}=\sin(x)\cdot \dfrac{1}{\cos(x)}\] Product rule gives \[f'(x)=\cos(x)\cdot \dfrac{1}{\cos(x)} + \sin(x)\cdot \left(\dfrac{1}{\cos(x)}\right)'=1+\sin(x)\cdot \left(\dfrac{1}{\cos(x)}\right)'\] We can calculate the derivative #\left(\dfrac{1}{\cos(x)}\right)'# by writing #\dfrac{1}{\cos(x)}# as the composition of #\dfrac{1}{x}# and #\cos(x)#. Applying the chain rule gives \[\left(\dfrac{1}{\cos(x)}\right)'= -\dfrac{1}{\cos(x)^2}\cdot (-\sin(x))=\dfrac{\sin(x)}{\cos(x)^2}\] We now conclude the proof: \[f'(x)=1 + \sin(x)\cdot \left(\dfrac{1}{\cos(x)}\right)'=1+\sin(x)\cdot \dfrac{\sin(x)}{\cos(x)^2}=1+\dfrac{\sin(x)^2}{\cos(x)^2}=1+\tan(x)^2\]