52–54. verifying derivative formulas verify the following derivative formulas using the quotient rule. 52…

Explanation:

Step1: Recall the Quotient Rule

The Quotient Rule states that if $y = \frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$.

Step2: Rewrite $\cot x$ and find its derivative

We know that $\cot x=\frac{\cos x}{\sin x}$. Let $u = \cos x$ and $v=\sin x$. Then $u'=-\sin x$ and $v'=\cos x$. Using the Quotient Rule: [ \begin{align*} \frac{d}{dx}(\cot x)&=\frac{-\sin x\cdot\sin x-\cos x\cdot\cos x}{\sin^{2}x}\ &=\frac{-\left(\sin^{2}x + \cos^{2}x\right)}{\sin^{2}x} \end{align*} ] Since $\sin^{2}x+\cos^{2}x = 1$, we have $\frac{d}{dx}(\cot x)=-\csc^{2}x$.

Step3: Rewrite $\sec x$ and find its derivative

We know that $\sec x=\frac{1}{\cos x}$. Let $u = 1$ and $v=\cos x$. Then $u'=0$ and $v'=-\sin x$. Using the Quotient Rule: [ \begin{align*} \frac{d}{dx}(\sec x)&=\frac{0\cdot\cos x-1\cdot(-\sin x)}{\cos^{2}x}\ &=\frac{\sin x}{\cos^{2}x}\ &=\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}\ &=\sec x\tan x \end{align*} ]

Answer:

The derivative $\frac{d}{dx}(\cot x)=-\csc^{2}x$ and $\frac{d}{dx}(\sec x)=\sec x\tan x$ are verified using the Quotient Rule.