How do you find the derivative of $(x+cosx)/tanx$ ?

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Answer 1 · 2017-04-30T16:56:40

$(df)/(dx)=(tanx-tanxsinx-xsec^2x-secx)/tan^2x$

We can use Quotient rule, which states if $f(x)=(g(x))/(h(x))$

then $(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2$

hence as $f(x)=(x+cosx)/tanx$

$(df)/(dx)=(tanx xx (1-sinx)-sec^2x(x+cosx))/tan^2x$

= $(tanx-tanxsinx-xsec^2x-secx)/tan^2x$

How do you find the derivative of (x+cosx)/tanx(x+cosx)/tanx?