Asked by Kim
Use implicit differentiation to find the slope of the tangent line to the curve at the point (4,1)
-1x^2 - 4xy + 3y^3 = -29
I differentiated both sides and solved for dy/dx and got 2x-4y/9y^2-4x. Then I plugged in X and Y and got -4/7
but when i enter my answer it's wrong. What is the right answer? Where did I go wrong?
-1x^2 - 4xy + 3y^3 = -29
I differentiated both sides and solved for dy/dx and got 2x-4y/9y^2-4x. Then I plugged in X and Y and got -4/7
but when i enter my answer it's wrong. What is the right answer? Where did I go wrong?
Answers
Answered by
Reiny
I got (2x+4y)/(9y^2-4x) for dy/dx
I bet your error was in the derivative of
-4xy
which is -4x(dy/dx) - 4y
check your signs.
I bet your error was in the derivative of
-4xy
which is -4x(dy/dx) - 4y
check your signs.
Answered by
Kim
Thanks a lot, it was the signs that was wrong.
Answered by
drwls
Differentiate both sides of the equation with respect to x, with y being a function of x.
-2x -4y -4x*dy/dx +9y^2*dy/dx = 0
dy/dx = [2x +4y]/[9y^2 -4x]
When x=4 and y = 1, I get
dy/dx = (8 + 4) /(9 - 16) = -12/7
-2x -4y -4x*dy/dx +9y^2*dy/dx = 0
dy/dx = [2x +4y]/[9y^2 -4x]
When x=4 and y = 1, I get
dy/dx = (8 + 4) /(9 - 16) = -12/7
Answered by
Kim
Thanks :)
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