I’m trying to get a derivative of $\sec(x)$ with respect to x.
The correct derivative is $\dfrac{\sin(x)}{\cos(x)^2}$, though this is not match the value that I was trying to solve.
Since $\sec(x) =\dfrac{1}{ \cos(x)}, \sec(x)’ =\bigg(\dfrac{1}{\cos(x)}\bigg)’ = \dfrac{-1}{\cos(x)^2}$
But when I applied a random value, such as $\frac{\pi}{3}$, to the equation, the first (correct) derivative gets me 3.46 while the second one gets me -4.
I wonder what I’m making a mistake on. I think the problem is I used power rule instead of quotient rule, but am not sure why power rule here gives me the wrong value. In fact, I tried to solve $x^{-1}$ using both quotient rule and power rule, both got me $\dfrac{-1}{x^2}$.
Best Answer
You have to apply the chain rule: $(f(g(x))^{'}=f^{'}(g(x))\cdot g^{'}(x)$ In your case it is
$\left(\sec(x)\right)^{'} = \left(\cos^{-1}(x)\right)^{'}=-1\cdot \cos^{-2}(x)\cdot (-\sin(x))=\tan(x)\cdot \sec(x)$