How do you find the derivatives of $y = {(sin θ)}^{tan θ}$ $y=(sintheta)^tantheta$ by logarithmic differentiation?

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How do you find the derivatives of $y = {(sin θ)}^{tan θ}$ $y=(sintheta)^tantheta$ by logarithmic differentiation?

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Answer 1 · 2017-02-10T10:08:27

$\frac{d y}{d x} = {(sin θ)}^{tan θ} ({sec}^{2} θ ln sin θ + 1)$ $(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)$

Explanation:

usinf logarithmic differentiation

$y = {(sin θ)}^{tan θ}$ $y=(sintheta)^(tantheta)$

$ln y = {ln (sin θ)}^{tan θ}$ $lny=ln(sintheta)^(tantheta)$

by laws of logs

$ln y = tan θ ln (sin θ)$ $lny=tanthetaln(sintheta)$

now differentiate $w r t x$ $wrt" " x$

$\frac{d}{d x} (ln y = tan θ ln (sin θ))$ $d/dx(lny=tanthetaln(sintheta))$

$R H S$ $RHS" "$ will need the product rule

$\frac{1}{y} \frac{d y}{d x} = {sec}^{2} θ ln sin θ + tan θ \frac{1}{sin θ} \times cos θ$ $1/y(dy)/(dx)=sec^2thetalnsintheta+tantheta1/sinthetaxxcostheta$

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

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

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

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

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How do you find the derivatives of $y = {(sin θ)}^{tan θ}$ $y=(sintheta)^tantheta$ by logarithmic differentiation?

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Explanation:

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How do you find the derivatives of $y = {(sin θ)}^{tan θ}$ $y=(sintheta)^tantheta$ by logarithmic differentiation?