# another question! - differentiationWatch

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Thread starter 13 years ago
#1
the question is to differentiate (x-4)/2(x-2)^3/2

so far i have:

2(x-2)^3/2 - (x-4)3(x-2)^1/2 / {2(x-2)^3/2}^2

=

2(x-2)^3/2 - 3(x-4)(x-2)^1/2 / 4(x-2)^3

but the next bit in the answer is:

(x-2)^1/2{2(x-2)-3(x-4)} / 4(x-2)^3

but im not suure how they got the above...so would appreciate it if ne1 could help!
0
13 years ago
#2
f(x)=(x-4)/2(x-2)^3/2

u=x-4, du/dx = 1
v=2(x-2)^3/2, dv/dx = 3(x-2)^½

Quotient rule:
f '(x)=[v(du/dx)-u(dv/dx)]/v²
f '(x)=[2(x-2)^3/2-3(x-4)(x-2)^½]/[2(x-2)^3/2]²
f '(x)=(x-2)^½[2(x-2)-3(x-4)]/[2²(x-2)³]
f '(x)=(x-2)^1/2[2(x-2)-3(x-4)]/4(x-2)³
0
Thread starter 13 years ago
#3
but how did u get:

f '(x)=(x-2)^1/2[2(x-2)-3(x-4)]/4(x-2)³

:s
0
13 years ago
#4
Take (x-2)^½ out as a common factor and tidy up. (Realising that (X^3/2)/(X^½) = X.)
0
Thread starter 13 years ago
#5
ohhh! thank you!!!
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