Question

Find the slope of the tangent line to the curve
\sqrt{1x +4y} + \sqrt{4xy} = 17
at the point ( 7,5 ).

Specifically, I'm having trouble with the \sqrt{4xy} since it requires both product and chain rules (not sure which to do first and how).

Answer 1

glad that you specified,
whenever in doubt, try substitution, u= 4xy
\((\sqrt{4xy})'=(\sqrt{u})'=1/2\sqrt{u}\times (4xy)'\)
now product rule for (4xy)'
don't forget to resubstitute back for 'u'

Answer 2

sry, i don't understand, this is what I have so far:
(1/2)(x+4y)^(-1/2) * (1+4(dy/dx)) + (1/2)[(4y)(4x(dy/dx))]^(-1/2)

Answer 3

regarding 'which to do first '
do, chain rule first,
so, there'll be 1/2 (4xy)^(-1/2) [....here product rule for (4xy)'....]

Answer 4

(1/2)(x+4y)^(-1/2) * (1+4(dy/dx)) + (1/2)(xy)^(-1/2) * (4y)(4x(dy/dx))
Here is what I have after those steps, I'm still uncertain what to do next

Answer 5

product rule for (4xy)' = 4 [y + x dy/dx] right ?
so, 2nd term = 1/2 (4xy)^(-1/2) 4[y + x dy/dx]
got this ?
1st term is just fine.

Answer 6

more specifically, i mean im not sure what to do with the first terms, since i need to bring all the y primes on one side and the rest to the other (so i can plug in points (7,5). Im just not sure about how to deal with (1/2)(x+4y)^(-1/2) and the steps afterwards.

Answer 7

you can plug in 7,5 before isolating y' also,
this will simplify isolating y' afterwards.