The chain rule and the composite function.

18.01 Calculus Jason Starr Fall 2005 Lecture 4. September 15, 2005Homework. No new problems.Practice Problems. Course Reader: 1F11 1F66 1F77 1F881. Product rule example. For u = √3x + 1, what is u�(x)? Since u · u =3x + 1, (u · u)� = (3x + 1)� = 3. By the product rule, (u · u)� = u� u� = 2uu�. Thus solving, · u + u · u�(x) = 3/(2u) = x −1/2/2 .3(3+ 1)2. The derivative of un . From above, (u2)� equals 2uu�. By a similar computation, (u3)� equals 3u2u�. This suggests a pattern, d(un) n−1 du = nu . dx dx 18.01 Calculus Jason Starr Fall 2005 This can be proved by induction on n. For n =1, 2 and 3, it was checked. Let n be a particular integer (for instance, 70119209472933054321). For that integer, suppose the result is known, d(un) n−1 du = nu . dx dx The goal is to prove the result for n + 1, that is, n+1)d(udu = (n + 1)u n . dx dx Let v = un . Then un+1 equals uv. So, by the product rule, n+1)d(ud(uv) du dv == v + u. dx dxdx dx Plugging in v = un, this is, n+1)d(udu d(un) = dx · (u n) + u. dx dx By the induction hypothesis, d(un)/dx equals nun−1(du/dx). Plugging in, dud(un+1) du ) + u(nu n−1 ). dx = dx · (u ndx This simplfies to, n+1)d(udu du du n = u + nu n = (n + 1)u n . dx dxdx dx Thus, the result for n + 1 follows from the result for n. By induction, the result holds for every n. an3. The derivative of xa , a a fraction. Let a be a fraction m/n and let u(x) be x. Then uequals xm . Thus, d(un) d(xm) = ,dx dx which equals mxm−1 . By the above, d(un)/dx equals nun−1(du/dx). Thus, du n−1 m−1 nu = mx . dx Solving for du/dx, du mxm−1 mxm−1 == . n(xm/n)n−1dx nun−1 b)cOne of the basic rules of exponents is that (aequals abc . Thus the denominator n(xm/n)n−1 m−m/nequals nxm/n(n−1), which equals nx. Thus, du mxm−1 m m−1 m/n−m == xx. nxm−m/ndx n · � � 18.01 Calculus Jason Starr Fall 2005 cAnother basic rule of exponents is that ab aequals ab+c . Thus, · du m (m−1)+(m/n−m) m m/n−1 = x= x. dxn n Remembering that m/n is just a, and u(x) is xa, this finally gives, d(xa) a−1 = ax . dx 4. The chain rule. Let y be a function of x, y = f (x), and let u be a function of y, u = g(y). Then u is a function of x, u = g(f (x)). This function is a composite function, and is denoted by, (g ◦ f )(x) = g(f (x)). What is the derivative of a composite function? The claim is that, (g ◦ f )�(x) = g�(f (x)) · f �(x). This is often easier to remember in the form, du dudy = . dx dy · dx This also suggests the proof, Δu Δu Δy(g ◦ f )�(x0) =lim =lim Δx0 Δx Δx0 Δy · Δx, →→where y0 equals f (x0), u0 equals g(y0) = g(f (x0)), Δy equals f (x0 + Δx) − f (x0) = f (x0 + Δx) − y0, and Δu equals g(y0 +Δy)−g(y0) = g(f (x0 +Δx))−g(f (x0)). So long as Δy is nonzero, the fraction in the limit is defined. And, as Δx approaches 0, also Δy approaches 0. Thus the limit breaks up as, Δu Δy(g ◦ f )�(x0) = lim lim Δy→0 Δy · Δx0 Δx = g�(y0) · f �(x0). →Thus (g ◦ f )�(x0) equals g�(f (x0))f �(x0). Example. Let y(x) equals 1 + x2, and let u(y) equal 1/y = y−1 . Then y�(x) = 0 + 2x =2x and u�(y) = −y−2 . Thus, by the chain rule, d 1 = −2 1(2x) = dx 1 + x2 y−2x (1+x2)2 . 5. Implicit differentiation. This method has already been used many times. Given a function y(x) satisfying some equation involving both x and y, formally differentiate each side of the equation with respect to x and then try to solve for y�.