Calculus-Slope of tangent line to curve

[chucknorrisfish]

Find the slope of the tangent line to the curve:
sqrt(4x+y) + sqrt(xy) = sqrt(39) + sqrt(56) at the point (8,7)

Here's what i've got....

.5(4x+y)^-.5(4+(dy/dx)) + .5(xy)^-.5 (y+x(dy/dx)) = 0

4+(dy/dx) + y+x(dy/dx) = 0
2sqrt(4x+y) 2sqrt(xy)


(dy/dx)(1/ 2sqrt(4x+y) + x/(2sqrt(xy))) = -4/2sqrt(4x+y) - y/ 2sqrt(xy)

These are the answers i've have gotten: .787963 , 12.94034929, and 18.62857182

am i doing this right?

 

[galactus]

I get -1.28210333793

The line isn't going to have 3 different slopes at the same point.

 

[stapel]

These are the answers i've have gotten: .787963 , 12.94034929, and 18.62857182

How did you get three slopes for one tangent line? :shock:

For this exercise, you must differentiate implicitly:

. . . . .sqrt(4x + y) + sqrt(xy) = sqrt(39) + sqrt(56)

. . . . .(1/2)(1/sqrt(4x + y))(4 + dy/dx) + (1/2)(1/sqrt(xy))(y + x(dy/dx)) = 0

. . . . .2/sqrt(4x + y) + (1/2)(1/sqrt(4x + y))(dy/dx) + (y/2)(1/sqrt(xy)) + (x/2)(1/sqrt(xy))(dy/dx) = 0

Isolate the "dy/dx" terms:

. . . . .(1/2)(1/sqrt(4x + y))(dy/dx) + (x/2)(1/sqrt(xy))(dy/dx) = -2/sqrt(4x + y) - (y/2)(1/sqrt(xy))

. . . . .(dy/dx)[1/(2sqrt(4x + y)) + x/(2sqrt(xy))] = -2/sqrt(4x + y) - y/(2sqrt(xy))

. . . . .dy/dx = [-2/sqrt(4x + y) - y/(2sqrt(xy))] / [1/(2sqrt(4x + y)) + x/(2sqrt(xy))]

Then plug in the given values of x and y. Simplify to find the value of dy/dx.

Note: You may be expected to provide the answer in "exact" form, with the radicals, rather than a decimal approximation.

Eliz.

 

[chucknorrisfish]

I didn't get three slopes, what i meant was that each time i tried to solve it i got a different answer, so i was probably doing some sort of algebraical error. But thanks guys, i got it now.