Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a function, $f(x) = \sec x + \csc x$ on the interval $x \in (0, \pi/2)$.

The derivative of it is, $f'(x) = \csc^2(x) \sec^2(x) \left(\sin^3 x + \cos^3 x\right)$

Of course, when I tried to solve for $f'(x) = 0$, I realize that $f'(x)=0$ does not exist.

However, the question insists that the there is a minimum. Am I doing something wrong here?

share|improve this question
I don't mean to give you too much of the answer, but Wolfram Alpha is a really powerful tool when you need more insight about these kinds of problems: wolframalpha.com/input/?i=f%28x%29+%3D+sec+x+%2B+csc+x It appears your derivative is incorrect. –  Jonathan Rich Mar 14 '13 at 17:38
Hint: Derivative is: $tan(x) sec(x)-cot(x) csc(x)$, $x = \pi/4$ –  Amzoti Mar 14 '13 at 17:39
I expanded from tan(x)sec(x)−cot(x)csc(x) to (sin x)/(cos^2 x) + (cos x)/(sin^2 x). I am sorry for not formatting properly. –  0xFF Mar 14 '13 at 17:40
You lost a $-$ sign –  Robert Israel Mar 14 '13 at 17:42
(sin x)/(cos^2 x) - (cos x)/(sin^2 x) => (sin^3x-cos^3x)/(cos^2 x)(sin^2 x) => (sec^2)(csc^2 x)(sin^3x-cos^3x) Am I wrong with my expansion? –  0xFF Mar 14 '13 at 17:45

3 Answers 3

up vote 3 down vote accepted

We should have $$\begin{align}f'(x) &= \frac{\sin x}{\cos^2 x}-\frac{\cos x}{\sin^2x}\\ &= \frac1{\cos^2x\sin^2x}\bigl(\sin^3x-\cos^3x\bigr)\\ &= \frac1{\cos^2x\sin^2x}(\sin x-\cos x)\bigl(\sin^2x+\sin x\cos x+\cos^2x\bigr)\\ &= \frac1{\cos^2x\sin^2x}(\sin x-\cos x)\left(1+\frac12\sin 2x\right).\end{align}$$ The fraction and the right-most factor at the end can't be zero for any real $x$. Hence, $f'(x)=0$ if and only if $\sin x=\cos x$. Where does that happen?

share|improve this answer

The derivative is $0$ at $x=\frac{\pi}{4}$ as the derivative goes from $-$ to $+$ you really have a minimum.

The correct derivative is $$-\cot(x) \csc(x) + \sec(x) \tan(x)$$

enter image description here

share|improve this answer

No roots of $f'(x)$ implies there is no local extrema. But on a closed interval, the minimum could also occur at a boundary point...

share|improve this answer
That would be true. But in this case there is a zero of $f'(x)$ in the interval, and $f(x) \to +\infty$ as you approach the boundary points. –  Robert Israel Mar 14 '13 at 17:40

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.