1
$\begingroup$

According to wolfram the derivative of this is zero. This derivative though is part of a larger problem and I don't think it is possible for it to be zero. According to my own calculation the answer is $\sec(x) \tan(x) + 2\sin(x)$
Is this correct?

$\endgroup$
3
$\begingroup$

You wrote it incorrectly and your answer is correct. What wolfram calculated was $(\sec (x))(-2\cos (x))$ which is actually $$\frac{1}{\cos (x)}\cdot\frac{-2\cos (x)}{1} =-2$$ And the derivative of $-2$ is $0$. When writing that equation in wolfram you need to pay careful attention to the brackets you use. As you would have now realized, $$\sec(x)-2\cos(x) \ne (\sec (x))(-2\cos (x))$$ There's no need for the extra brackets on the outside...This indicates multiplying the two terms.

$\endgroup$
0
$\begingroup$

This derivative cannot possibly be zero. What you have is correct.

$\endgroup$
0
$\begingroup$

$y=\sec(x)-2\cos(x)\\\implies \dfrac{dy}{dx}=\tan(x)\sec(x)+2\sin(x)\neq 0$.

Indeed you are right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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