Question

Kalkulus diferensial, mohon bantuanya teman teman.
disuruh mencari dy/dx . terimakasih ^^

Answer 1

Saya hanya bisa jawab no. 4
gini nih..
misal :
u =x-2 --> u'=1
v=x^(3/4) --> v'=(3/4)x^(-1/4)

dy/dx = (u'v-uv')/v^2
dy/dx = ((1)(x^(3/4))-((x-2)((3/4)x^(-1/4)) dibagi (x^(3/4))^2
setelah dihitung menghasilkan..
((1/4)x^(3/4)+(6/4)x^(-1/4))/x^(6/4)
segitu jawabannya, saya tidak bisa menyederhanakan lebih lanjut, udah menthok.

maaf ya kalo salh... senang bisa membantu =)

Answer 2

A)

x^2.y^3 = e^(3x)

u = x^2
u' = 2x
v = y^3
v' = 3y^2.y'

2x.y^3 + x^2.3y^2.y' = 3e^(3x)
3(xy)^2.y' = 3e^(3x) - 2x.y^3
y' = e^(3x)/(xy)^2 - (2/3)(y/x)
dy/dx = e^(3x)/(xy)^2 - (2/3)(y/x)

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b)

x^2.cos(xy) = e^(2x)

u = x^2
u' = 2x
v = cos(xy)
v' = (y + xy')(-sin(xy))

(2x)(cos(xy)) + x^2(y + xy')(-sin(xy)) = 2e^(2x)
(y + xy') = (2e^(2x) - 2x.cos(xy))/(-x^2.sin(xy))
xy' = (2e^(2x) - 2x.cos(xy))/(-x^2.sin(xy)) - y
y' = (2e^(2x) - 2x.cos(xy))/(-x^3.sin(xy)) - y/x

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c)

e^(x/2y) = ln y

u = e^(x/(2y))
u' = (1/(2y) - (xy')/(2y^2))e^(x/(2y))

(1/(2y) - (xy')/(2y^2))e^(x/(2y)) = y'/y
e^(x/2y)/(2y) - xy'e^(x/2y)/(2y^2) - y'/y = 0
- xy'e^(x/2y)/(2y^2) - y'/y = -e^(x/2y)/2y
xy'e^(x/2y)/(2y^2) + y'/y = e^(x/2y)/2y
y'(x.e^(x/2y)/2y^2 + 1/y) = e^(x/2y)/2y
y' =  e^(x/2y)/2y / ((x.e^(x/2y)/2y^2 + 1/y))

==========

d)

y = (x - 2)/(x^(3/4))
y = x^(1/4) - 2.x^(-3/4)
dy/dx = -(3/4)x^(-3/4) + 2.(7/4).x^(-7/4)
dy/dx = -(3/4)x^(-3/4) + (7/2)x^(-7/4)

CMIIW