Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a book where are shown steps to derivate $\tan(x)$ incompletely:

$D(\dfrac{\sin x}{\cos x})$

$= \dfrac{\cos x D(\sin x) - \sin x D(\cos x)}{\cos^2 x}$

$= \dfrac{\cos x \cos x + \sin x \sin x}{\cos^2 x}$

$= \dfrac {1}{\cos^2 x} = \sec^2 x$

Can you explain me how to get to step 2?

share|cite|improve this question
up vote 2 down vote accepted

You may want to take a look here: It's called the quotient rule for derivatives.

$\textbf{Added.}$ Well, if you don't like that then use the standard definition of derivative which is $$ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ Using this with $f(x)=\tan{x}$ we have

\begin{align*} f'(x) &= \lim_{h \to 0} \frac{\tan(x+h)-\tan{x}}{h} \\ &= \lim_{h \to 0} \frac{\frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin{x}}{\cos{x}}}{h} \\ &=\lim_{h \to 0} \frac{ \sin(x+h)\cdot \cos{x} - \cos(x+h)\cdot\sin{x}}{h} \times \frac{1}{\cos(x+h)\cdot \cos{x}} \\ &=\lim_{h \to 0} \frac{\sin(x+h -x)}{h} \cdot \lim_{h \to 0} \frac{1}{\cos(x+h)\cdot \cos{x}} \qquad \Bigl[ \small\because \sin(A-B) = \sin{A}\cdot \cos{B} - \cos{A}\cdot \sin{B} \Bigr] \\ &= \frac{1}{\cos^{2}{x}} = \sec^{2}{x} \qquad\qquad \qquad \Bigl[ \small\because \lim_{x \to 0} \frac{\sin{x}}{x} =1 \Bigr] \end{align*}

$\textbf{Added 2.}$ Or if you are aware of the product rule then you can also do like this: $$ (uv)' = u'v + uv'$$ and note that

  • $u=\sin{x}$

  • $v = \frac{1}{\cos{x}} = \sec{x}$ and it's derivative $v' == \sec{x} \cdot \tan{x}$

share|cite|improve this answer

we have $D(\frac{u}{v})=\frac{D(u)v-D(v)u}{v^2}$.now $u=sinx$ and $v=cosx$ so $D(\dfrac{\sin x}{\cos x})=1+tg^2x$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.