I found this problem while studying online. When I take the derivative of $\sin x^{\cos x}$ using the rule: $$(a^x)'=a^x\ln a,$$ and using the chainrule I get: $$\sin x^{\cos x} \ln \sin x\cdot(-\sin x).$$ However in the page I found this problem on, the solution is listed as: $$\sin x^{\cos x}\left(\frac {\cos^2x} {\sin x}-\sin x\ln(\sin x)\right)$$ I don't believe my solution was wrong but I'd like to know how to get to the solution that's on the page. Thanks in advance :-)
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See How would I differentiate $\sin{x}^{\cos{x}}?$ – Sil Apr 18 '17 at 20:00
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Sorry, I searched this problem by the term derivative instead of differentiation and it displayed no results. – tot11 Apr 18 '17 at 21:02
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The main problem is that the rule you are using assumes $a$ is a constant. Whilst your problem is most definitely not. $$ u(x) = f(x)^{g(x)} = \mathrm{e}^{g\ln f} $$ which means $$ u'(x) = \left(g'(x) \ln f(x) + \frac{g(x)}{f(x)}f'(x)\right)f(x)^{g(x)} $$
Chinny84
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Note your answer is incorrect, since the formula you quoted requires $a$ to be a constant
Let $y=\sin x^{\cos x}$
$$\ln y = \cos x\ln(\sin x)$$
$$\frac{1}{y} \frac{dy}{dx} = -\sin x\ln(\sin x)+\cos x\cdot\frac{\cos x}{\sin x}$$
$$\frac{dy}{dx} = \sin x^{\cos x}\bigg(\frac{\cos^2x}{\sin x} -\sin x\ln(\sin x)\bigg) $$
mrnovice
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I suggest you start here:
$(\sin x)^{cos x} = e^{\cos x \ln (\sin x)}$
As you differentiate don't lose track of the product rule and the chain rule.
$\frac d{dx} e^{\cos x \ln \sin x} = e^{\cos x \ln \sin x} \frac d{dx} (\cos x\ln (\sin x))\\ e^{\cos x \ln \sin x} (-\sin x \ln (\sin x) + \frac {cos x}{\sin x}(\frac d{dx} \sin x)\\ e^{\cos x \ln \sin x} (-\sin x \ln (\sin x) + \frac {cos^2 x}{\sin x})$
And finally we can simplfy $e^{\cos x \ln \sin x}$ back again
$(\sin x)^{cos x} (-\sin x \ln (\sin x) + \frac {cos^2 x}{\sin x})$
Doug M
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$${ \left( { \sin { x } }^{ \cos { x } } \right) }^{ \prime }={ \left( { e }^{ \cos { x } \ln { \sin { x } } } \right) }^{ \prime }={ e }^{ \cos { x } \ln { \sin { x } } }\left( -\sin { x } \ln { \left( \sin { x } \right) +\frac { \cos ^{ 2 }{ x } }{ \sin { x } } } \right) \\ $$
haqnatural
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Write the function as $$ y=e^{\cos x \ln(\sin x)} $$
so
$$ y'=e^{\cos x \ln(\sin x)}\left[-\sin x \ln(\sin x) +\frac{\cos x}{\sin x} \cos x \right] $$
Emilio Novati
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