Implicit differentiation might come up a few weeks into the semester. \u00a0It’s a nice technique that enables the student to take derivatives of functions not solved for y.<\/p>\n
Example:<\/strong><\/p>\n Find the derivative of xy^2 -siny = 11<\/p>\n Solution:<\/strong> With implicit differentiation, we first assume that y is some function of x which we don’t know. We might imagine y=f(x). What we are trying to find is y’, which might also be referred to as f'(x).<\/p>\n Following the point of view that y=f(x), we can rewrite the equation with f(x) instead of y:<\/p>\n x(f(x))^2 – sin(f(x)) = 11<\/p>\n Now, we take the derivative from left to right on each side. x(f(x))^2 requires the product rule (uv)’ = u’v + uv’<\/p>\n (x(f(x))^2)’= 1(f(x))^2 + x(2f(x)f'(x))<\/p>\n Notice the chain rule in the second part: (f(x)^2)’=2f(x)f'(x). First, we take the derivative of the outer function with the power rule. Then, we multiply it by the derivative of f(x) itself, f'(x).<\/p>\n Next we take the derivative of -sin(f(x)), once again invoking the chain rule:<\/p>\n (-sin(f(x)))’=-cos(f(x))f'(x)<\/p>\n On the right side, the derivative of 11 is 0. Writing the derivative of each term in the equation, we get<\/p>\n (f(x))^2 + 2xf(x)f'(x) – cos(f(x))f'(x) = 0<\/p>\n What we are really trying to find is f'(x). We need to isolate it using algebra. First, we get all the terms that don’t include f'(x) onto the other side:<\/p>\n 2xf(x)f'(x) – cos(f(x))f'(x) = -(f(x))^2<\/p>\n Next, we factor out f'(x) from the left as a common factor:<\/p>\n f'(x)[2xf(x) – cos(f(x))] = -(f(x))^2<\/p>\n Finally, we divide both sides by 2xf(x) – cos(f(x)) to isolate f'(x):<\/p>\n f'(x)=-((f(x))^2)\/(2xf(x) – cos(f(x)))<\/p>\n Since the original assumption was that y=f(x), it follows that f'(x) is the derivative. If desired, we can substitute y and y’ back into the solution so it matches the original context:<\/p>\n y’=(-y^2)\/(2xy-cosy)<\/p>\n HTH:)<\/p>\n Source<\/em>:<\/p>\n Larson and Hostetler. Calculus: Part One<\/em>. Toronto: D. C. Heath and Company, 1989. <\/p>\n