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 $-\sin(x)$.
Best Answer
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.