Math, asked by PragyaTbia, 1 year ago

Differentiate the function w.r.t.x:
\rm \frac{1}{cosec\ x+\cot x}


amannoufel: where are you even getting these points.............
amannoufel: you have nt even answered a question

Answers

Answered by Anonymous
0
HOPE IT HELPS U ✌️✌️
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Answered by hukam0685
0
We know that such form of expression can solve by U/V method of differentiation

 \frac{d}{dx} \bigg(\frac{U}{V}\bigg ) = \frac{V \frac{dU}{dx} - U \frac{dV}{dx} }{ {V}^{2} } \\ \\ here \: U = 1 \\ \\ V = cosec\:x+cot \: x \\\\
\frac{d}{dx} \bigg(\frac{1}{cosec\:x+cot \: x} \bigg) = \frac{(cosec\:x+cot \: x) \frac{d(1)}{dx} -( 1 )\frac{d(cosec\:x+cot \: x)}{dx} }{( {cosec\:x+cot \: x})^{2} } \\ \\

\frac{d}{dx} \bigg(\frac{1}{cosec\:x+cot \: x} \bigg) = \frac{(cosec\:x+cot \: x)(0) -( 1 )(-cosec\:x\:cot\:x-{cosec}^{2} x) }{( {cosec\:x+cot \: x})^{2} } \\ \\ \frac{d}{dx} \bigg(\frac{1}{cosec\:x+cot \: x} \bigg)= \frac{cosec\:x(cosec\:x+cot\:x)}{( {cosec\:x+cot \: x})^{2} } \\ \\ \frac{d}{dx} \bigg(\frac{1}{cosec\:x+cot \: x} \bigg)= \frac{cosec\:x}{( {cosec\:x+cot \: x})} \\ \\

Hope it helps you.
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