## Calculus: Early Transcendentals 8th Edition

$\dfrac{d}{dx}(\csc x)=-\csc x\cot x$
$\dfrac{d}{dx}(\csc x)$ We know that $\csc x=\dfrac{1}{\sin x}$, so we make that substitution and differentiate applying the quotient rule: $\dfrac{d}{dx}(\dfrac{1}{\sin x})=\dfrac{(\sin x)(1)'-(1)(\sin x)'}{(\sin x)^{2}}=...$ $...=\dfrac{(\sin x)(0)-(1)(\cos x)}{\sin^{2}x}=\dfrac{-\cos x}{\sin^{2}x}$ This is the derivative. of $\csc x$. Since we have two factors of sine in the denominator, let's write this expression as a product of two fractions: $\dfrac{-\cos x}{\sin^{2}x}=(\dfrac{1}{\sin x})(-\dfrac{\cos x}{\sin x})$ Since $\dfrac{1}{\sin x}=\csc x$ and $\dfrac{\cos x}{\sin x}=\cot x$, we finally get: $\dfrac{d}{dx}(\csc x)=-\csc x\cot x$