1 mark for review which of the following correctly shows the derivation of $\frac{d}{dx}(cot x)$? (a)…

1 mark for review which of the following correctly shows the derivation of $\frac{d}{dx}(cot x)$? (a) $\frac{d}{dx}(cot x)=\frac{d}{dx}(\frac{1}{\tan x})=\frac{1}{\frac{d}{dx}(\tan x)}=\frac{-1}{sec^{2}x}$ (b) $\frac{d}{dx}(cot x)=\frac{d}{dx}(\frac{1}{\tan x})=\frac{1}{\frac{d}{dx}(\tan x)}=\frac{1}{sec^{2}x}$ (c) $\frac{d}{dx}(cot x)=\frac{d}{dx}(\frac{1}{\tan x})=\frac{\tan x\frac{d}{dx}(1)-1cdot\frac{d}{dx}(\tan x)}{\tan^{2}x}=\frac{(\tan x)cdot0 - sec^{2}x}{\tan^{2}x}=-\frac{sec^{2}x}{\tan^{2}x}$ (d) $\frac{d}{dx}(cot x)=\frac{d}{dx}(\frac{1}{\tan x})=\frac{\tan x\frac{d}{dx}(1)+1cdot\frac{d}{dx}(\tan x)}{\tan^{2}x}=\frac{\tan xcdot0+sec^{2}x}{\tan^{2}x}=\frac{sec^{2}x}{\tan^{2}x}$

1 mark for review which of the following correctly shows the derivation of $\frac{d}{dx}(cot x)$? (a) $\frac{d}{dx}(cot x)=\frac{d}{dx}(\frac{1}{\tan x})=\frac{1}{\frac{d}{dx}(\tan x)}=\frac{-1}{sec^{2}x}$ (b) $\frac{d}{dx}(cot x)=\frac{d}{dx}(\frac{1}{\tan x})=\frac{1}{\frac{d}{dx}(\tan x)}=\frac{1}{sec^{2}x}$ (c) $\frac{d}{dx}(cot x)=\frac{d}{dx}(\frac{1}{\tan x})=\frac{\tan x\frac{d}{dx}(1)-1cdot\frac{d}{dx}(\tan x)}{\tan^{2}x}=\frac{(\tan x)cdot0 - sec^{2}x}{\tan^{2}x}=-\frac{sec^{2}x}{\tan^{2}x}$ (d) $\frac{d}{dx}(cot x)=\frac{d}{dx}(\frac{1}{\tan x})=\frac{\tan x\frac{d}{dx}(1)+1cdot\frac{d}{dx}(\tan x)}{\tan^{2}x}=\frac{\tan xcdot0+sec^{2}x}{\tan^{2}x}=\frac{sec^{2}x}{\tan^{2}x}$

Answer

Explanation:

Step1: Recall cotangent - tangent relation

Since $\cot x=\frac{1}{\tan x}$, we use the 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 $v=\tan x$.

Step2: Find derivatives of $u$ and $v$

We know that $\frac{d}{dx}(1) = 0$ and $\frac{d}{dx}(\tan x)=\sec^{2}x$.

Step3: Apply the quotient - rule

$\frac{d}{dx}(\cot x)=\frac{d}{dx}\left(\frac{1}{\tan x}\right)=\frac{\tan x\frac{d}{dx}(1)-1\cdot\frac{d}{dx}(\tan x)}{\tan^{2}x}=\frac{\tan x\cdot0 - \sec^{2}x}{\tan^{2}x}=-\frac{\sec^{2}x}{\tan^{2}x}$.

Answer:

C. $\frac{d}{dx}(\cot x)=\frac{d}{dx}\left(\frac{1}{\tan x}\right)=\frac{\tan x\frac{d}{dx}(1)-1\cdot\frac{d}{dx}(\tan x)}{\tan^{2}x}=\frac{(\tan x)\cdot0-\sec^{2}x}{\tan^{2}x}=-\frac{\sec^{2}x}{\tan^{2}x}$