I have to find the slope of the tangent line to the curve at the point (2, 8):
√4x+4y + √2xy = 12
Or also can be written as:
(4x+4y)^(1/2) + (2xy)^(1/2) = 12
I tried differentating got got:
(d/dx)[(4x+4y)^(1/2)] = (1/2)*(4x+4y)^(-1/2)*[4+4(dy/dx)]
(d/dx)[(2xy)^(1/2)] = (1/2)*(2xy)^(-1/2)*[2y+2x*(dy/dx)]
then,
(2+2*(dy/dx))/(4x+4y)^1/2 + (4+x*dy/dx)/(2xy)^(1/2)
I then plugged in numbers and ended up with (4+4sqrt5)/((-2sqrt5)-4). However that answer is wrong. Can anyone give a clue as to what i did wrong?
Hello, FrustratedCollegeKid!
It would help if you told us the (alleged) right answer.
Also, there must be typo in the problem.
. . The point (2, 8) is not on the graph.
Find the slope of the tangent line to the curve at the point (2, 8):
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We have: .
Then: .
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As a sidenote, when required to get the value of dy/dx at a given point, I find it much easier NOT to first make dy/dx the subject. Rather, I'd substitute the point into the line marked * and then simplify and make dy/dx the subject. Less algebra, less chance of error, less time taken.