What is the derivative of this function $f(x)=1/sinx$ ?

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Answer 1 · 2018-05-28T16:56:02

The answer is $=-cosx/sin^2x$

The function is

$f(x)=1/sinx=(sinx)^-1$

The derivative is

$f'(x)=((sinx)^-1)'=(-(sinx)^-2)*cosx$

$=-cosx/sin^2x$

Answer 2 · 2018-05-28T17:02:07

$-cotxcscx$ .

We will use the Quotient Rule (QR) to find $f'(x)$ .

QR : $f(x)=g(x)/(h(x))," then, "f'(x)={h(x)g'(x)-g(x)h'(x)}/[h(x)]^2$ .

$:. f(x)=1/sinx rArr f'(x)={sinx*d/dx(1)-1*d/dx(sinx)}/[sinx]^2$ ,

$={sinx*0-1*cosx}/sin^2x$ ,

$=-cosx/sin^2x$ ,

$=-cosx/sinx*1/sinx$ ,

$=-cotxcscx$ .

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