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sin y + cos x = x + 1, which gives rise to a (x, y) = (0, 0) is a solution to that equation, and there is a line tangent to the curve at (0, 0). We examined the equation in the plane. Note that curve is nice enough, curve if the Even though we are not able to write down a formula we know how to diery in terms of x near (0, 0) (in fact, y = sin−1 (1 + x − cos x) works for entiate for y between −π/2 π/2), and we can use the method of implicit dierentiation to have the chain rule tell us the slope of the tangent line at (0, 0). sin y + cos x = x + 1 ⇒ ⇒ d d (sin y + cos x) = (x + 1) dx dx d d d d sin y + cos x = x+ 1 dx dx dx dx ⇒ ⇒ (cos y) (outer function is d sin y − sin x = 1 + 0 dx d y − sin x = 1 dx f (x) = sin x, ⇒ by the chain rule inner function is g (x) = y near (0, 0) ) 1 + sin x d y= dx cos y Therefore, dy 1 + sin 0 = =1 dx (x,y)=(0,0) cos 0 Since the slope of the tangent line through line is y − 0 = 1 (x − 0), or y = x. (0, 0) is 1, the equation of the The curve together with this tangent line looks like the following: 1
