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Numerade Educator

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Problem 19 Easy Difficulty

Prove that $ \frac {d}{dx}$ (cot $ x $) = $ - csc^2 x. $

Answer

$\frac{d}{d x}(\cot x)=\frac{d}{d x}\left(\frac{\cos x}{\sin x}\right)=\frac{(\sin x)(-\sin x)-(\cos x)(\cos x)}{\sin ^{2} x}=-\frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x}=-\frac{1}{\sin ^{2} x}=-\csc ^{2} x$

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Video Transcript

he It's clear, so enumerated here. So we're gonna do a proof We're gonna start off with our code tangent. So de over de ex coat engine of hacks. Another way to right. This is D over D x of co sign over sign, which is equal to sign when we apply the quotient. Cool de over de ex co sign minus co sign D over the access sign well over sign square. This becomes equal to sign times Negative sign dryness Co sign turns co signed over signed square, which is equal to negative Signed Square Plus Co signed square over a sign square. This becomes equal to negative one for trig identities over sign at square, which is equal to negative co Sika Square.