Please help derives the following
f (x) = e ^ x [(x-9) / (x-7)] ^ 1 / 2

Question

Please help derives the following
 f (x) = e ^ x [(x-9) / (x-7)] ^ 1 / 2

amistre64 · Answer

is this a product of "e" and "x-9"/"x-7"?

amistre64 · Answer

$$f (x) = (e ^ x)\left(\cfrac{(x-9)}{(x-7)}\right) ^ {1 / 2}$$

anonymous · Answer

Yuk...

amistre64 · Answer

[rl]' = r'l + rl'   ; my right and left get transposed, but i think it 
                         looks better that way  :)

r = e^x ;  r'=e^x

l=$\left(\cfrac{x-9}{x-7}\right)^{^1/_2}$  ;  l' = a bit complicated :)

anonymous · Answer

x-7 = u^2

anonymous · Answer

i did this super fast so could be wrong, but i get:

e^x * (x - 7)^(-3/2) * (x - 8)^2 / (x - 9)^(1/2)

amistre64 · Answer

[t b^(-1)] = t b' + t' b

t = x-9;  t' = 1

b = (x-7)^-1 ; b'= (-x+7)^-2

$\cfrac{x-9}{(7-x)^2}$ + $\cfrac{1}{x-7}$

looks to the the derivative of the fractional part; b ut then incorporate the sqrt into it ... ugh

amistre64 · Answer

i got b' mixed up :)

anonymous · Answer

what kind of sadist is asking this question?

amistre64 · Answer

yeah lol.   practice problems are spose to build confidence :)

anonymous · Answer

e^(u^2+7)(u^2-2)^(1/2)