hey guys i have a little problem i cant get my head around
i undertsand that
but this question
thats (x-3)^0.5 which is root (i dont know how to root in the maths tags
but the answer i keep getting is
the real answer is
hey guys i have a little problem i cant get my head around
i undertsand that
but this question
thats (x-3)^0.5 which is root (i dont know how to root in the maths tags
but the answer i keep getting is
the real answer is
hey mate,
as you stated you have an equation of the form
y = u(x)v(x) for which you want to find the derivative which as you stated is solved using the product rule,
i.e.,
y'(x) =u'(x)v(x) + u(x)v'(x)
here,
(i) u(x) = (x-9) --> u'(x) = 1
(ii) v(x) = (x-3)^(1/2) --> v'(x) = (1/2)*(x-3)^(-1/2) (I'm assuming you know the chain rule - please advise me if you dont and I'll be more than happy to explain)
Therefore
y'(x) = (1)*(x-3)^(1/2) + (x-9)*(1/2)*(x-3)^(-1/2) = (2(x-3) + (x-9))/(2*(x-3)^(1/2))
= (1/2)*(3x - 15)/(x-3)^(1/2)
Hope this helps,
Let me know if you require any further assistance,
David
hey mate,
not a problem - just in reference to the (1/2) component I believe you were having troubles with, as you would know it composed in the derivative of,
v(x) = (x-3)^(1/2)
Here we employ the Chain rule, i.e. let w = x-3 then
v = w^(1/2) --> v'(w) = (1/2)w^(-1/2) (i.e. d/dx (x^n) = nx^(n-1) )
and w'(x) = 1
Therefore by the Chain Rule,
v'(x) = v'(w)*w'(x) or v'(x) = (dv/dw)*(dw/dx)
Thus,
v'(x) = (1/2)w^(-1/2) * (1)
Recall w = x-3
Therefore
v'(x) = (1/2)(x-3)^(-1/2)
Hope this helps,
Regards,
David
ps - are you good with your logarithms? there is another technique which makes alot of these nasty derivates simpler to handle.