Before I attempt this .....
I am curious why you would put a coefficient of 1 in front of the variables, such as in
(1x + 2y) and (1xy)
At this stage of Calculus, they would certainly be understood and not needed.
I am tempted to guess you meant
(1/x + 2y) and √(1/xy)
please clarify.
Find the slope of the tangent line to the curve √(1x+2y) + √(1xy) = 8.24 at the point (2,8)?
I know you have to use implicit differentiation, but the radicals keep making me mess up algebraically. Is the changing the radicals to exponents the fastest way? please show me the steps thanks
(1x)^(1/2) + (2y)^(1/2) + (1xy)^(1/2)=8.24
4 answers
no it's actually 1x and 1xy. It's not division.
√(1x+2y) + √(1xy) = 8.24
changing the radicals to exponents is really the only way, explicitly or implicitly, but
√(1x+2y) ≠ √(1x) + √(2y) !!!
Anyway, we have, using implicit differentiation (and deleting those useless 1 coefficients),
√(x+2y) + √(xy) = 8.24
1/2√(x+2y) (1+2y') + 1/2√(xy) (y+xy') = 0
1/2√(x+2y) + y/2√(xy) + y'/√(x+2y) + xy'/2√(xy) = 0
y'(1/√(x+2y) + x/2√(xy)) = -(1/2√(x+2y) + y/2√(xy))
y' =
-(y√(x+2y) + √(xy)) / (x√(x+2y) + 2√(xy))
changing the radicals to exponents is really the only way, explicitly or implicitly, but
√(1x+2y) ≠ √(1x) + √(2y) !!!
Anyway, we have, using implicit differentiation (and deleting those useless 1 coefficients),
√(x+2y) + √(xy) = 8.24
1/2√(x+2y) (1+2y') + 1/2√(xy) (y+xy') = 0
1/2√(x+2y) + y/2√(xy) + y'/√(x+2y) + xy'/2√(xy) = 0
y'(1/√(x+2y) + x/2√(xy)) = -(1/2√(x+2y) + y/2√(xy))
y' =
-(y√(x+2y) + √(xy)) / (x√(x+2y) + 2√(xy))
thanks
2 answers