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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)

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take the derivative and set it 0
im stuck on getting the deriviative

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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?
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
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
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
still coming up incorrect :/

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