Find the slope of the tangent line to the curve -4x^2-4xy-3y^3 at the point (6,2)
same process as the last problem... how far did you get?
so we find the derivitive of 4x^2-4xy-3y^3?
you use implicit differentiation, yes... to find y'
how would i set it up?
not sure what you're asking... did you use implicit differentiation? what result did you get?
i didn't, im confused on how to use implicit differentiation
you treat y as some function of x. we don't really need to know what exactly that function of x actually is... we don't care in fact. It seems weird but it works. Go through term by term and differentiate with respect to x... eg.: first term (-4x^2 ) ' pretty easy: -8x second term (-4xy) ' now we have a product of x and y use the product rule: -4 ( x'y +y'x) x' is just 1 y' is y' (we don't know explicitly what function of x y is... doesn't matter, we just leave it as y') so (-4xy)' = -4(y +y'x)
Isn't there a missing term in the equation of the curve?
third term... use the chain rule (since y is a function of x) try it: ( -3y^3) ' =...?
-9y^2?
pretty good, but y' "pops out", because of the chain rule... so -9y^2 * y'
ohh,
anyway... if your original equation was something like -4x^2-4xy-3y^3 =C we now have the result of differentiating that equation with respect to x: -8x -4(y +y'x) -9y^2 * y' = 0
and now we solve for y prime?
all that's left to do is solve for y', which is just a matter of algebra :)
yep:)
im getting -14/3 but its coming out incorrect
you can imagine y is some well known function of x and take the derivative and you'll see that it works... eg.: if y = sinx (-3(sin(x))^3 )' = -9(sin(x))^2 * cos(x) { -9 y ^2 * y' ) ! (I left out a 'prime' in the original, fixt)
what did you get as the answer what i got s showing as incorrect
nevermind, i found what i did wrong. Thanks for the help!
cool.:)
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