What is the value of the constant c so that x-y=c is the equation of the line tangent to the graph f(x)=In(Sin(x)) where 0<x<pi I know that line x-y=c has a slope of 1, so therefore f'x = 1 I do not what to do next
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Originally Posted by 05holtel What is the value of the constant c so that x-y=c is the equation of the line tangent to the graph f(x)=In(Sin(x)) where 0<x<pi I know that line x-y=c has a slope of 1, so therefore f'x = 1 I do not what to do next Take the derivative: $\displaystyle \frac{d}{dx}\left[\ln{(\sin{x})}\right]=\cot{x}\Rightarrow\cot{x}=1\Rightarrow\cos{x}=\si n{x}\Rightarrow{x}=\frac{\pi}{4}$
Last edited by VonNemo19; Jul 26th 2009 at 05:27 PM. Reason: Made a boo-boo.
Originally Posted by VonNemo19 Take the derivative: $\displaystyle \frac{d}{dx}\left[\ln{(\sin{x})}\right]=\cot{x}\Rightarrow\cot{x}=1\Rightarrow\cos{x}=\si n{x}\Rightarrow{x}=\frac{\sqrt{2}}{2}$ correction ... $\displaystyle x = \frac{\pi}{4}$
Originally Posted by skeeter correction ... $\displaystyle x = \frac{\pi}{4}$ Thank you. Got a little ahead of myself...
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