Question

which of the following correctly shows the derivation of $\frac{d}{dx}(sec x)$?
(a) $\frac{d}{dx}(sec x)=\frac{d}{dx}(\frac{1}{cos x})=\frac{1}{\frac{d}{dx}(cos x)}=\frac{1}{sin x}$
(b) $\frac{d}{dx}(sec x)=\frac{d}{dx}(\frac{1}{cos x})=\frac{1}{\frac{d}{dx}(cos x)}=\frac{1}{-sin x}$
(c) $\frac{d}{dx}(sec x)=\frac{d}{dx}(\frac{1}{cos x})=\frac{\frac{d}{dx}(1)cdotcos x - 1cdot\frac{d}{dx}(cos x)}{(cos x)^2}=\frac{0cdotcos x - 1cdot(-sin x)}{(cos x)^2}$
(d) $\frac{d}{dx}(sec x)=\frac{d}{dx}(\frac{1}{cos x})=\frac{\frac{d}{dx}(1)cdotcos x+1cdot\frac{d}{dx}(cos x)}{(cos x)^2}=\frac{0cdotcos x + 1cdot(-sin x)}{(cos x)^2}$

Explanation:

Step1: Recall secant - cosine relationship

We know that $\sec x=\frac{1}{\cos x}$.

Step2: Apply quotient - rule for differentiation

The quotient - rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 1$ and $v=\cos x$. So, $u^\prime=\frac{d}{dx}(1)=0$ and $v^\prime=\frac{d}{dx}(\cos x)=-\sin x$.
Then $\frac{d}{dx}(\sec x)=\frac{d}{dx}(\frac{1}{\cos x})=\frac{\frac{d}{dx}(1)\cdot\cos x - 1\cdot\frac{d}{dx}(\cos x)}{\cos^{2}x}=\frac{0\cdot\cos x-1\cdot(-\sin x)}{\cos^{2}x}$.

Answer:

C. $\frac{d}{dx}(\sec x)=\frac{d}{dx}(\frac{1}{\cos x})=\frac{\frac{d}{dx}(1)\cos x - 1\cdot\frac{d}{dx}(\cos x)}{(\cos x)^{2}}=\frac{0\cdot\cos x - 1\cdot(-\sin x)}{(\cos x)^{2}}$