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Find dy/dx of y = (2x+3)/(x-5)

2007-08-28 19:29:20 · 4 answers · asked by ppp 1 in Science & Mathematics ➔ Mathematics

4 answers

dy/dx = [(x-5) (2) - (2x+3) * 1] / (x-5)^2
dy/dx = (2x-10-2x-3)/ (x-5)^2
dy/dx = -13 / (x-5)^2

2007-08-28 19:55:12 · answer #1 · answered by gab BB 6 · 1⤊ 0⤋

Using the quotient rule:
(u/v)' = (u'v-uv')/v^2

dy/dx = [(2x+3)'(x-5)-(2x+3)(x-5)']/(x-5)^2
= [2(x-5)-(2x+3)]/(x-5)^2
= (2x-10-2x-3)/(x-5)^2
= -13/(x-5)^2

2007-08-28 19:38:54 · answer #2 · answered by Anonymous · 0⤊ 0⤋

y = (2x+3) / (x-5)
y = (2x+3) * (x-5)^-1
y = 2(x-5)^-1 -(2x+3)(x-5)^-2

2007-08-28 19:41:11 · answer #3 · answered by 037 G 6 · 0⤊ 0⤋

Substitute s=(x-5). Then y=(2s-7)/s. dy/ds =d(2s-7)/ds(1/s) +(2s-7)d(1/s)/dx.

Then Use dy/dx =dy/ds(ds/dx).

2007-08-28 19:42:46 · answer #4 · answered by stvenryn 4 · 0⤊ 0⤋