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

Question

12121 views around the world

You can reuse this answer
Creative Commons License

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#.