Spivak asks us to evaluate $$\dfrac{d}{dx}\{(\sin x)^{\cos x} + (\cos x) ^{\sin x}\}$$ by logarithmic differentiation. Does he mean for us to evaluate each term separately (which seems to turn out to be cumbersome), or is there something else I'm missing?
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$\begingroup$ You mean evaluate the derivative? $\endgroup$– Nathaniel BubisFeb 27, 2014 at 6:56
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$\begingroup$ A logarithmic derivative sometimes (usually?) means $$\lim_{\Delta x \rightarrow 0} \log_{\Delta x}\left(\frac{f(x + \Delta x)}{f(x)}\right)$$ the inverse of the geometric integral. $\endgroup$– DanielVFeb 27, 2014 at 7:18
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$\begingroup$ $\dfrac{d}{dx}\ln f(x)=\dfrac{f'(x)}{f(x)}$ $\endgroup$– LucianFeb 27, 2014 at 7:18
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2$\begingroup$ I do not see what Spivak means by $evaluate$ in the case of this problem. Is it $evaluate$ $the$ $derivative$ $of$ $(\sin x)^{\cos x} + (\cos x) ^{\sin x}$ ? If this is the case, I suggest you modify the title of the post. $\endgroup$– Claude LeiboviciFeb 27, 2014 at 9:41
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$\begingroup$ Where in Spivak is this exercse? $\endgroup$– DonAntonioFeb 27, 2014 at 11:26
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2 Answers
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Hint. Take $f(x)=(\sin x)^{\cos x}$. You try to differentiate $f(x)+f(\pi/2 -x)$. Use logarithmic differentiation to derivative $f(x)$.
answered Feb 27, 2014 at 7:26
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Let $f(x) = u+v$ where $u = (\sin x)^{\cos x}$ and $v = (\cos x)^{\sin x}$
As we know, If $f(x) = u+v, f'(x) = \dfrac{du}{dx} + \dfrac{dv}{dx}$
Consider $u = (\sin x)^{\cos x}$
$\ln u = \cos x \ln(\sin x)$
Differentiating, $\dfrac{1}{u}\dfrac{du}{dx} = \cos x \dfrac{\cos x}{\sin x} - \sin x \ln(\sin x)$ --- (A). by product rule
Now do similarly for $\dfrac{dv}{dx}$ and find $\dfrac{du}{dx}+\dfrac{dv}{dx} = f'(x)$
Edit: You will have to multiply (A) by $u$ to get $\dfrac{du}{dx}$
