MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 30 July 2014

Lecture 4

Last time we pointed out that there are different kinds of derivatives: Derivative with respect tox Derivative with respect tog f ddx dfdx f ddg dfdg This one satisfies This one satisfies dxdx = 1 d(y+z)dx = dydx+ dzdx d(cy)dx = cdydx, ifcis a constant d(yz)dx = ydzdx+ dydxz dgdg = 1 d(y+z)dg = dydg+ dzdg d(cy)dg = cdydg, ifcis a constant d(yz)dz = ydzdg+ dydgz What is the relation between dfdx and dfdg?? ddg ddx dg0dg = d1dg=0 dg0dx = d1dx=0 dgdg = 1 dgdx = dgdx dg2dg = d(g·g)dg = gdgdg+ dgdgg = g+g=2g dg2dx = d(g·g)dx = gdgdx+ dgdx·g = 2gdgdx dg3dg = d(g2·g)dg = g2dgdg+ dg2dgg = g2+2g·g=3g2 dg3dx = d(g2·g)dx = g2dgdx+ dg2dxg = g2dgdx+ 2gdgdxg = g2dgdx+ 2g2dgdx dg4dg = d(g4·g)dg = g3dgdg+ dg3dg·g = g3+3g2·g = 4g3 dg4dx = d(g3·g)dx = g3dgdx+ dg3dxg = g3dgdx+ 3g2dgdxg = 4g3dgdx dg6342dg = 6342g6341 dg6342dx = 6342g6341dgdx d(3g2+2g+7)dg = d(3g2)dg+ d(2g)dg+ d7dg = 3dg2dg+ 2dgdg+0 = 3·2g+2·1 = 6g+2 d(3g2+2g+7)dx = d(3g2)dx+ d(2g)dx+ d7dx = 3dg2dx+ 2dgdx+ d7dx = 3·2gdgdx+ 2dgdx+0 = 6gdgdx+ 2dgdx = (6g+2)dgdx If f is any function, then we have the chain rule dfdx= dfdg dgdx, i.e. f ddx dgdx dgdx f ddg dfdg.

Find dydx when y=(2x-5)2.

If g=2x-5 then y=g2. dydx= dydg dgdx = dg2dg dgdx=2g d(2x-5)dx =2g(2) = 4(2x-5).

Find ddx when y=(3x-4)3.

If g=3x-4 then y=g3. dydx= dydg dgdx = dg3dg dgdx= 3g2dgdx= 3(3x-4) d(3x-4)dx = 3(3x-4)·3= 27x-36.

Find dydx when y=(2x-5)2(3x-4)3. dydx = d(2x-5)2(3x-4)3dx = (2x-5)2 d(3x-4)3dx+ d(2x-5)2dx (3x-4)3 = (2x-5)2 3(3x-4)2· d(3x-4)dx +(3x-4)3 2(2x-5) d(2x-5)dx = 3(2x-5)2 (3x-4)2·3+2 (2x-5) (3x-4)3·2 = (2x-5)(3x-4)2 (9(2x-5)+4(3x-4)) = (2x-5) (3x-4)2 (30x-61).

Find dydx when y=(x-3x-4)2. d(x-3x-4)2dx = 2(x-3x-4) d(x-3x-4)dx = 2(x-3x-4) d((x-3)(x-4)-1)dx = 2(x-3x-4) ( (x-3)d(x-4)-1dx+ d(x-3)dx(x-4)-1 ) = 2(x-3x-4) ( (x-3)(-1) (x-4)-2 d(x-4)dx+ 1·(x-4)-1 ) = 2(x-3x-4) ( -(x-3)(x-4)2 ·1+1x-4 ) = 2(x-3x-4) ( -x+3(x-4)2 +x-4(x-4)2 ) = 2(x-3x-4) (-1)(x-4)2 = -2x+6(x-4)3

Find dxmndx. d(xmn)ndx= dxmdx=mxm-1. On the other hand d(xmn)ndx= n(xmn)n-1 dxmndx. So mxm-1=n (xmn)n-1 dxmndx and we can solve for dxmndx. dxmndx = mxm-1 n(xmn)n-1 = mxm-1 n(xmn)n(xmn)-1 = mxm-1 mxm1xmn = mnx-1xmn =mnxmn=1. So dxmndx= mnxmn-1.

Find dydx when y=x1-2x. dydx = dx1-2xdx = dx(1-2x)-1dx = dx((1-2x)12)-1dx = dx(1-2x)-12dx = xd(1-2x)-12dx+ dxdx(1-2x)-12 = x(-12)(1-2x)-32 d(1-2x)dx+1· 11-2x = -x2(1-2x)32 ·(-2)+1(1-2x)12 = x+1-2x (1-2x)32 = 1-x(1-2x)32.

Notes and References

These are a typed copy of Lecture 4 from a series of handwritten lecture notes for the class MATH 221 given on September 13, 2000.

page history