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I wanted to prove for myself, using chain rule instead of quotient rule, that $\sec'(x)$ is $\sec(x)\tan(x)$ but I got $\tan(x)$. That's how I did it:\begin{align*} \sec(x) = 1/\cos(x) & \Rightarrow y = \frac{1}{\cos(x)} = \cos(x)^{-1}\\\\ & \Rightarrow y' = \frac{-1}{\cos(x)}(-\sin(x)) = \frac{\sin(x)}{\cos(x)} = \tan(x) \end{align*}

What did I do wrong?

If you are wondering how did I get that result, first I took derivative of the outward funcion lets say $y = u^{-1}$ than I took derivative of it so $(-1)uu'$, where $u'$ is derivative of the function so in my case $\cos(x)$ which is $-\sin(x)$.

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2 Answers

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You forgot to reduce $1$ in the power of the fraction.

You can simply solve for the derivative by doing:$$\begin{align*} y & = \sec{(x)}\\ \frac{\mathbb{d}}{\mathbb{d}x}y & = \frac{\mathbb{d}}{\mathbb{d}x}\left(\frac{1}{\cos{(x)}}\right) \end{align*}$$

Now consider $u = \cos(x)$, then the equation becomes, using the chain rule:$$\begin{align*} y' & = \left(\frac{\mathbb{d}}{\mathbb{d}u}\frac{1}{u}\right)\cdot u'\\ & = \left(\frac{\mathbb{d}}{\mathbb{d}u} u^{-1}\right)\cdot u'\\ & = -u^{-2}\cdot u'\\ & = -\frac{1}{u^2}\cdot u'\\ & = -\frac{-\sin(x)}{\cos(x)\cos(x)}\\ & = \frac{1}{\cos(x)}\cdot\frac{\sin(x)}{\cos(x)} \end{align*}$$$$\boxed{y' = \sec(x)\cdot\tan(x)}$$

Hopefully you can spot the reduction.

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You should know that $$\biggl(\frac1u\biggr)'=-\frac1{u^2}\cdot u',$$and more generally that$$\biggl(\frac1{u^n}\biggr)'=-\frac n{u^{n+1}}\cdot u'.$$

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