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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 $\dfrac{-1}{x^2}$.

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2 Answers 2

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You have to apply the chain rule: $(f(g(x))^{'}=f^{'}(g(x))\cdot g^{'}(x)$ In your case it is

$\left(\sec(x)\right)^{'} = \left(\cos^{-1}(x)\right)^{'}=-1\cdot \cos^{-2}(x)\cdot (-\sin(x))=\tan(x)\cdot \sec(x)$

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    $\begingroup$ Thanks! I seemed to have a mistake in not seeing cos(x)^(-1) as a composite function of cos(x) and x^(-1). $\endgroup$ Commented Oct 29, 2022 at 12:21
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Alternatively, you can use the reciprocal rule: if $f$ is differentiable at $x$, and $f(x)\neq0$, then $1/f$ is differentiable at $x$, and $$ \left(\frac1f\right)'\left(x\right)=-\frac{f'(x)}{f(x)^2} \, . $$ Applying this theorem in the case $f=\cos$, we see that $$ \sec'(x)=\left(\frac1\cos\right)'\left(x\right)=-\frac{\cos'(x)}{\cos^2(x)}=\frac{\sin(x)}{\cos^2(x)}=\frac{1}{\cos(x)}\cdot\frac{\sin(x)}{\cos(x)}=\sec(x)\tan(x) \, , $$ at all points where $\cos(x)\neq0$, i.e. whenever $x$ is not of the form $\pi/2+k\pi,k\in\mathbb Z$.

The reciprocal rule is a special case of the quotient rule, which states that under the appropriate conditions, $$ \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2} \, . $$

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