Question

‌‌How‌ ‌do‌ ‌you‌ ‌differentiate‌ ‌$f(x)=\csc‌ ‌x$‌ ‌using‌ ‌the‌ ‌quotient‌ ‌rule?

Answer 1

We‌ ‌are‌ ‌given‌ ‌an‌ ‌expression‌ ‌which‌ ‌we‌ ‌have‌ ‌to‌ ‌differentiate‌ ‌using‌ ‌the‌ ‌quotient‌ ‌rule.‌ ‌First,‌ ‌we‌ ‌will‌ ‌write‌ ‌the‌ ‌expression‌ ‌in‌ ‌simplified‌ ‌form‌ ‌and‌ ‌we‌ ‌get,‌ ‌$f(x)=\dfrac{1}{\sin‌ ‌x}$.‌ ‌We‌ ‌will‌ ‌then‌ ‌use‌ ‌the‌ ‌quotient‌ ‌rule‌ ‌and‌ ‌substitute‌ ‌the‌ ‌values‌ ‌in‌ ‌the‌ ‌formula‌ ‌as‌ ‌required.‌ ‌On‌ ‌solving‌ ‌further,‌ ‌we‌ ‌get‌ ‌the‌ ‌differentiation‌ ‌of‌ ‌the‌ ‌given‌ ‌expression.‌

Complete step-by-step solution:
According‌ ‌to‌ ‌the‌ ‌given‌ ‌question,‌ ‌we‌ ‌are‌ ‌given‌ ‌an‌ ‌expression‌ ‌which‌ ‌we‌ ‌have‌ ‌to‌ ‌differentiate‌ ‌using‌ ‌the‌ ‌
quotient‌ ‌rule.‌ ‌
Quotient‌ ‌rule‌ ‌is‌ ‌a‌ ‌rule‌ ‌for‌ ‌differentiation‌ ‌of‌ ‌function‌ ‌which‌ ‌involves‌ ‌ratio‌ ‌of‌ ‌two‌ ‌functions‌ ‌which‌ ‌are‌ ‌
differentiable.‌ ‌The‌ ‌formula‌ ‌of‌ ‌the‌ ‌quotient‌ ‌rule‌ ‌is‌ ‌as ‌follows.‌ ‌
$\dfrac{d}{dx}\left(‌ ‌\dfrac{u}{v}‌ ‌\right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}$‌ ‌
We‌ ‌will‌ ‌carry‌ ‌out‌ ‌the‌ ‌differentiation‌ ‌based‌ ‌on‌ ‌the‌ ‌above‌ ‌formula‌ ‌of‌ ‌quotient‌ ‌rule.‌ ‌
The‌ ‌given‌ ‌expression‌ ‌we‌ ‌have‌ ‌is,‌ ‌
$f(x)=\csc‌ ‌x$-----(1)‌ ‌
We‌ ‌can‌ ‌simplify‌ ‌it‌ ‌further‌ ‌and‌ ‌write‌ ‌the‌ ‌equation‌ ‌(1)‌ ‌as,‌ ‌
$f(x)=\dfrac{1}{\sin‌ ‌x}$-----(2)‌ ‌
Using‌ ‌the‌ ‌formula‌ ‌of‌ ‌quotient‌ ‌rule,‌ ‌we‌ ‌will‌ ‌substitute‌ ‌the‌ ‌values‌ ‌in‌ ‌it.‌ ‌And‌ ‌we‌ ‌have,‌ ‌

x)}{{{\sin‌ ‌}^{2}}x}$$-----(3)‌ ‌ We‌ ‌proceed‌ ‌on‌ ‌to‌ ‌solve‌ ‌the‌ ‌above‌ ‌expression‌ ‌to‌ ‌get‌ ‌the‌ ‌differentiation‌ ‌value,‌ ‌we‌ ‌have,‌ ‌ $$\Rightarrow‌ ‌\dfrac{d}{dx}\left(‌ ‌\dfrac{1}{\sin‌ ‌x}‌ ‌\right)=\dfrac{\sin‌ ‌x(0)-1(\cos‌ ‌x)}{{{\sin‌ ‌}^{2}}x}$$‌ ‌ And‌ ‌as‌ ‌we‌ ‌know‌ ‌that,‌ ‌differentiation‌ ‌of‌ ‌any‌ ‌constant‌ ‌gives‌ ‌us‌ ‌a‌ ‌value‌ ‌0,‌ ‌that‌ ‌is,‌ ‌$$\dfrac{d}{dx}(1)=0$$‌ ‌ and‌ ‌differentiation‌ ‌of‌ ‌sine‌ ‌function‌ ‌is‌ ‌cosine‌ ‌function,‌ ‌that‌ ‌is,‌ ‌$$\dfrac{d}{dx}(\sin‌ ‌x)=\cos‌ ‌x$$.‌ ‌ So,‌ ‌we‌ ‌get,‌ ‌ $$\Rightarrow‌ ‌\dfrac{d}{dx}\left(‌ ‌\dfrac{1}{\sin‌ ‌x}‌ ‌\right)=\dfrac{-(\cos‌ ‌x)}{{{\sin‌ ‌}^{2}}x}$$‌ ‌ We‌ ‌can‌ ‌further‌ ‌proceed‌ ‌as,‌ ‌ $$\Rightarrow‌ ‌\dfrac{d}{dx}\left(‌ ‌\dfrac{1}{\sin‌ ‌x}‌ ‌\right)=\dfrac{-\cos‌ ‌x}{\sin‌ ‌x}.\dfrac{1}{\sin‌ ‌ x}$$-----(4)‌ ‌ And‌ ‌as‌ ‌we‌ ‌know‌ ‌that,‌ ‌$$\dfrac{\cos‌ ‌x}{\sin‌ ‌x}=\cot‌ ‌x$$‌ ‌and‌ ‌$$\dfrac{1}{\sin‌ ‌x}=\csc‌ ‌x$$,‌ ‌using‌ ‌these‌ ‌in‌ ‌ the‌ ‌equation‌ ‌(4),‌ ‌we‌ ‌get,‌ ‌ $$\Rightarrow‌ ‌\dfrac{d}{dx}\left(‌ ‌\dfrac{1}{\sin‌ ‌x}‌ ‌\right)=-\cot‌ ‌x\csc‌ ‌x$$‌ ‌ Therefore,‌ ‌we‌ ‌get‌ ‌the‌ ‌differentiation‌ ‌of‌ ‌$f(x)=\csc‌ ‌x$‌ ‌as:‌ ‌ $$f'(x)=-\cot‌ ‌x\csc‌ ‌x$$.‌ ‌ **Note:** ‌The‌ ‌pre-requisite‌ ‌of‌ ‌using‌ ‌the‌ ‌quotient‌ ‌rule‌ ‌is‌ ‌that‌ ‌the‌ ‌given‌ ‌expression‌ ‌should‌ ‌be‌ ‌in‌ ‌the‌ ‌form‌ of‌ ‌fraction‌ ‌(with‌ ‌a‌ ‌numerator‌ ‌and‌ ‌numerator).‌ ‌And‌ ‌that‌ ‌is‌ ‌why‌ ‌wrote‌ ‌$$f(x)=\csc‌ ‌x=\dfrac{1}{\sin‌ ‌x}$$.‌ ‌The‌ ‌formula‌ ‌should‌ ‌be‌ ‌written‌ ‌correctly‌ ‌and‌ ‌the‌ ‌values‌ ‌should‌ ‌be‌ ‌substituted‌ ‌in‌ ‌the‌ ‌formula‌ ‌in‌ ‌different‌ ‌steps‌ ‌to‌ ‌minimize‌ ‌mistakes.‌