What is the derivative of $f (x) = \frac{1}{sin x}$ $f(x)=1/sinx$ ?

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Answer 1 · 2015-06-08T13:44:33

Since $f (x) = \frac{1}{sin (x)} = csc (x)$ $f(x)=1/(sin(x))=csc(x)$ , the answer can be written down from memorization that $f'(x)=-csc(x)cot(x)$ .

Alternatively, the Quotient Rule can be used: $f'(x)=\frac{sin(x)*0-1*cos(x)}{sin^{2}(x)}=-\frac{cos(x)}{sin^{2}(x)}$

$=-\frac{1}{sin(x)}*\frac{cos(x)}{sin(x)}=-csc(x)cot(x)$

What is the derivative of f(x)=1/sinxf(x)=1sinxf(x)=1/sinx?