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Find the slope of the tangent line to the curve sqrt(3x +3y) +sqrt(2xy) =11.5 at points (4,5)

Mathematics
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take the derivative and set it 0
im stuck on getting the deriviative
1/2(3x+3y)^(-1/2)(3+3y')+1/2(2xy)^(-1/2)(2y+2xy')=0

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Other answers:

sub x=4 and y=5 find y'=m
after found m use this formula and write the eq. y-y1=m(x-x1) y1=5 x1=4
\[\frac{1}{2}(3x+3y)^{-1/2}(3+3y')+\frac{1}2 (2xy)^{-1/2}(2y+2xy')=0\]
i dont think im getting the right answer what did you get as the answer?
\[y'=m=\frac{4\sqrt{3}-2\sqrt{10}}{2\sqrt{10}-5\sqrt{3}}\]
use your cal..
-13/50 approx.
its showing as incorrect what was the exact answer?
what are the options..
i dont have options, but the assignment is online, so it tells me if he answer is correct or not
\[y=\frac{-13}{50}x+\frac{53}{10}\]
or \[y-4=(\frac{4\sqrt{3}-2\sqrt{10}}{2\sqrt{10}-5\sqrt{3}})(x-5)\]
is it saying round your answer or exact
exact
\[y=\frac{4\sqrt{3}-2\sqrt{10}}{2\sqrt{10}-5\sqrt{3}}x-5\frac{4\sqrt{3}-2\sqrt{10}}{2\sqrt{10}-5\sqrt{3}}+4\]
let me double check it
wht would the slope be?
\[y=-\frac{15\sqrt{2}+2\sqrt{15}}{12 \sqrt{2}+2\sqrt{15}}x+5\frac{15\sqrt{2}+2\sqrt{15}}{12 \sqrt{2}+2\sqrt{15}}+4\]
slope is coeff of x
its sill coming up as incorrect
\[y=-\frac{\sqrt{10}+4\sqrt{3}}{\sqrt{10}+4\sqrt{3}}x+5\frac{\sqrt{10}+4\sqrt{3}}{\sqrt{10}+4\sqrt{3}}+4\]
still coming up incorrect :/
\[y=\frac{-12\sqrt{3}+15}{2\sqrt{3}+12}x+\frac{60\sqrt{3}-75}{2\sqrt{3}+12}+4\]

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