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: . 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 .
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:
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