$\\frac{d}{dx}(\\cot x) =$ \na $-\\tan x$ \nb $-\\csc^{2}x$ \nc $\\csc x$ \nd $\\sec^{2}x$
$\\frac{d}{dx}(\\cot x) =$ \na $-\\tan x$ \nb $-\\csc^{2}x$ \nc $\\csc x$ \nd $\\sec^{2}x$
Answer
Explanation:
Step1: Recall cotangent identity
$\cot x=\frac{\cos x}{\sin x}$
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, $u = \cos x$, $u^\prime=-\sin x$, $v=\sin x$, and $v^\prime=\cos x$. So, $\left(\cot x\right)^\prime=\frac{-\sin x\cdot\sin x-\cos x\cdot\cos x}{\sin^{2}x}$.
Step3: Simplify the expression
$\frac{-\sin^{2}x - \cos^{2}x}{\sin^{2}x}=\frac{-(\sin^{2}x+\cos^{2}x)}{\sin^{2}x}$. Since $\sin^{2}x+\cos^{2}x = 1$, we have $\frac{- 1}{\sin^{2}x}=-\csc^{2}x$.
Answer:
B. $-\csc^{2}x$