By Niels Walet

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5-18 Often we take a function of a function. In such a case, where y = f (g(x)) we put z = g(x), and find dy dy dz = . dx dz dx This rule is sometimes expressed in words as “the differative of the function, times the derivative of its argument”, and you may know it as dg(f (x)) = f (g(x))g (x). 1. 1: Find dy dx for y = cos(ln x). Solution: Put z = ln x so y = cos z, dy dy = , dx dz sin(ln x) dz 1 = − sin z = − dx x x . 2: Find dy dx for y = sin3 (2x − 1). 7 . 3: Given that x(t) = 5t2 m, find the velocity v(t) and the acceleration a(t).

7: 4 Evaluate I = 0 1 √ dx. 1+ x Solution: Put x = z 2 , dx/dz = 2z, dx = 2zdz. The limits change, x = 0 ⇒ z = 0, x = 4 ⇒ z = 2. 4 Integration by Parts L&T, 15,21-30 This is the integral equivalent to the differential of a product. Start with d(uv) dv du =u +v . dx dx dx Integrate both sides with respect to x, uv = u dv dx + dx v du dx dx . Now use (dv/dx)dx = dv and (du/dx)dx = du. Rearrange the terms, and find udv + vdu. u dv = uv − v du. 8: Evaluate I = xex dx. Solution: Put u = x and ex dx = dv.

21: Given the relation between current, voltage and resistance, I = V /R, with V = 250 V, R = 50 Ω, find the change in the current I if V changes by 1 V, and R by 1 Ω. Solution: We use the rule for small changes for partial derivatives, δI ≈ ∂I ∂I δV + δR . ∂V ∂R We find ∂I ∂V ∂I ∂R = = 1 , R −V . 08 A 50 50 50 10 25 Chapter 5 Integration This chapter should contain partially things you know -essentially the basis of integration- and quite a few new things that build on that, extending your knowledge of integrals and integration.