Answers – Advanced Differentiation

1)
y’ex + yex = 2xy + x2y’ – 7y – 7xy’
y’ex - x2y’ + 7xy’  = 2xy – 7y - yex
y’ = (2xy – 7y – yex)/(ex – x2 + 7x)

2)
-3/x2 – y’/4y2 = 0
y’/4y2 = -3/x2
y’ = -12y2/x2

3)
9(1/2)(y')(2/y)sin x + 9(ln y/2)cos x = extan 2y + ex(sec2 2y)2y'
y'(9/y)sin x - y'2ex(sec2 2y) = extan 2y - 9(ln y/2)cos x
y'[(9/y)sin x - 2ex(sec2 2y)] = extan 2y - 9(ln y/2)cos x
y' = [extan 2y - 9(ln y/2)cos x]/[(9/y)sin x - 2ex(sec2 2y)]
4)
ln y = 4xsin x ln (sin x)
y’/y = 4sin x ln (sin x) + 4xcos x ln (sin x) + 4xcos x
y’ = (sin x)4xln x[4sin x ln (sin x) + 4xcos x ln (sin x) + 4xcos x]
5)
y’ = ex + xex
6)
ln y = (4x + 2)ln (x2 – 6x)
y’/y = 4 ln (x2 – 6x) + (4x + 2)(2x – 6)/(x2 – 6x)
y’ = (x2 – 6x)4x + 2[4 ln (x2 – 6x) + (4x + 2)(2x – 6)/(x2 – 6x)]
7)
ln y = xln (2x4 + ex)
y’/y = ln (2x4 + ex) + x(8x3 + ex)/(2x4 + ex)
y’ = (2x4 + ex)x[ln (2x4 + ex) + x(8x3 + ex)/(2x4 + ex)]
8)
ln y = tan x ln (tan x)
y’/y = sec2 x ln (tan x) + sec2 x
y' = (tan x)tan x[sec2 x ln (tan x) + sec2 x]

9)
ln y = sin x ln (ln 4x)
y’/y = cos x ln (ln 4x) + (sin x)/(xln 4x)
y’ = ycos x ln (ln 4x) + y(sin x)/(xln 4x)
y’ = (ln 4x)sin xcos x ln (ln 4x) + (ln 4x)sin x(sin x)/(xln 4x)

 

Copyright 2001 Bruce
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