Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x)) y'=cos[(sinx)(cosx)] y'3π=cos[(sinx)(-1)] cos [(0)(-1)] cos(0)=1 y-0=1(x-3π) y= x-3πWhat's wrong with my answer?
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Originally Posted by hazecraze Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x)) y'=cos[(sinx)(cosx)] y'3π=cos[(sinx)(-1)] cos [(0)(-1)] cos(0)=1 y-0=1(x-3π) y= x-3πWhat's wrong with my answer? Your derivative's wrong: $\displaystyle (\sin({\sin x}))'=\cos({\sin x})\cos x$. Tonio
Originally Posted by hazecraze Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x)) y'=cos[(sinx)(cosx)] You wrote that wierd, so I'm not sure if you got it wrong, or just used bad notation. y'3π=cos[(sinx)(-1)] cos [(0)(-1)] cos(0)=1 y-0=1(x-3π) y= x-3πWhat's wrong with my answer? The derivative is $\displaystyle cos(sin(x))cos(x)$. $\displaystyle y'(3\pi)=cos(sin(3\pi))cos(3\pi)=cos(0)(-1)=-1$
Originally Posted by hazecraze Find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x)) y'=cos[(sinx)(cosx)] <<<<<<<<< here. This line should read: y' = cos(sin(x)) * cos(x) y'3π=cos[(sinx)(-1)] cos [(0)(-1)] cos(0)=1 y-0=1(x-3π) y= x-3πWhat's wrong with my answer? The equation of the tangent line becomes now: $\displaystyle y = -x + 3\pi$
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