How do you find the derivatives of $y=(sintheta)^tantheta$ by logarithmic differentiation?

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Answer 1 · 2017-02-10T10:08:27

$(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)$

usinf logarithmic differentiation

$y=(sintheta)^(tantheta)$

$lny=ln(sintheta)^(tantheta)$

by laws of logs

$lny=tanthetaln(sintheta)$

now differentiate $wrt" " x$

$d/dx(lny=tanthetaln(sintheta))$

$RHS" "$ will need the product rule

$1/y(dy)/(dx)=sec^2thetalnsintheta+tantheta1/sinthetaxxcostheta$

$1/y(dy)/(dx)=sec^2thetalnsintheta+cancel((sintheta/costhetaxxcostheta/sintheta))^(=1)$

$1/y(dy)/(dx)=sec^2thetalnsintheta+1$

$(dy)/(dx)=y(sec^2thetalnsintheta+1)$

$(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)$

How do you find the derivatives of y=(sintheta)^tanthetay=(sintheta)^tantheta by logarithmic differentiation?