find d^2y/dx^2 for y=(1-x)/(2-x)

Question

amistre64 · Answer

find the second derivative ... right?

anonymous · Answer

yes

anonymous · Answer

try the quotient rule

amistre64 · Answer

then i would either use a product rule with a ^-1; or the quotient rule

anonymous · Answer

ok thanks so much

amistre64 · Answer

youd have to do it twice tho to get to the second one

anonymous · Answer

ok

amistre64 · Answer

$$\frac{t}{b}=tb^{-1}$$
$$D_x[tb^{-1}]=-tb^{-2}+t'b^{-1}$$
$$D_{x}^{2}[tb^{-1}]=D_x[-tb^{-2}+t'b^{-1}]$$
$$D_x[-tb^{-2}+b^{-1}]=2tb^{-3}-t'b^{-2}-t'b^{-2}+t''b^{-1}$$

witha any luck I kept it all in order lol

amistre64 · Answer

t=(1-x)
t' = -1
t'' = 0

b=(2-x)

$$2(1-x)(2-x)^{-3}+(2-x)^{-2}+(2-x)^{-2}+0(b^{-1})$$
$$2(1-x)(2-x)^{-3}+2(2-x)^{-2}$$
$$2(2-x)^{-3}\left((1-x)+(2-x)\right)$$
$$\frac{2(3)}{(2-x)^{3}}$$

the wolf should be able to check that tho

amistre64 · Answer

almost kept it straight :) http://www.wolframalpha.com/input/?i=second+derivative+t%28x%29%28b%28x%29%29%5E%28-1%29 2 ------- (x-2)^3 http://www.wolframalpha.com/input/?i=second+derivative+%281-x%29%2F%282-x%29