Find the slope of the tangent line to the curve
√4x+1y + √2xy = 6.06 @ the point (3,1)

I got this (-1-sqrt(2) sqrt(x) y)/(sqrt(x) (1+sqrt(2) x)), then I plugged in the values 3 and 1 the answer is -0.379877994777674

But for some reason I still got it wrong I guess not
sure

Any help?

Question

Find the slope of the tangent line to the curve 
√4x+1y  + √2xy = 6.06 @ the point (3,1) 

I got this (-1-sqrt(2) sqrt(x) y)/(sqrt(x) (1+sqrt(2) x)), then I plugged in the values 3 and 1 the answer is -0.379877994777674

But for some reason I still got it wrong I guess not 
sure 

Any help?

amistre64 · Answer

when more than one digit appears in the sqrt, its best to use the () to indicate its extent.

amistre64 · Answer

im assuming its 

√(4x+1y)  + √(2xy) = 6.06

anonymous · Answer

Yes....that's right

anonymous · Answer

Sorry

amistre64 · Answer

√(4x+1y)  + √(2xy) = 6.06

  (4x'+y')         2x'y+2xy'
--------- +  --------- = 0
2√(4x+1y)       2√(2xy)

amistre64 · Answer

4                        y'                 2y              2xy'
--------- +  --------- +---------+---------= 0
2√(4x+1y)    2√(4x+1y)       2√(2xy)      2√(2xy)

now we can gather the y's and solve

anonymous · Answer

so just plug in the values or we still have to do the y right?