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

1 Answer

Shwetank Mauria

Apr 30, 2017

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

Explanation:

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#

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