How do you find the derivative of #y=tan^ntheta#?

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Answer 1 · 2017-12-23T13:53:45

#(dy)/(d theta)=ntan^(n-1) thetasec^2 theta#

we need the chain rule

#(dy)/(d theta)=(dy)/(du)(du)/(d theta)#

#y=tan^n theta#

#u=tan theta=>(du)/(d theta)=sec^2 theta#

#:.y=u^n=>(dy)/(du)=n u^(n-1)#

#(dy)/(dx)=n u^(n-1)xxsec^2 theta#

#(dy)/(dx)=ntan^(n-1) thetasec^2 theta#