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

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

$y=tan^n theta$

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

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

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

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

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