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How to find dy/dx, if y=ln(8x^2+9y^2) ?

Answer 1

Isabella Ackerman

#dy/dx=(16x)/(8x^2+9y^2-18y)#

#y=ln(8x^2+9y^2)#

#dy/dx=(16x)/(8x^2+9y^2)+(18y)/ (8x^2+9y^2) dy/dx#

#dy/dx -(18y)/ (8x^2+9y^2) dy/dx=(16x)/(8x^2+9y^2)#

#dy/dx(1-(18y)/ (8x^2+9y^2) )=(16x)/(8x^2+9y^2)#

#dy/dx((8x^2+9y^2-18y)/ (8x^2+9y^2) )=(16x)/(8x^2+9y^2)#

#dy/dx=((16x)/(8x^2+9y^2))/((8x^2+9y^2-18y)/ (8x^2+9y^2) )#

#dy/dx=(16x)/(8x^2+9y^2-18y)#

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Answer 2

Ezra Adams

To find dy/dx, differentiate y with respect to x using the chain rule:

dy/dx = (d(ln(u))/du) * (du/dx)

Where u = 8x^2 + 9y^2

Now differentiate y = ln(u) and u = 8x^2 + 9y^2 separately:

dy/du = 1/u du/dx = d(8x^2 + 9y^2)/dx

Now differentiate u with respect to x:

du/dx = 16x + 18y(dy/dx)

Now substitute dy/du and du/dx back into the chain rule formula:

dy/dx = (1/u) * (16x + 18y(dy/dx))

Now solve for dy/dx:

dy/dx - 18y(dy/dx)/u = 16x/u

dy/dx(1 - 18y/u) = 16x/u

dy/dx = (16x/u) / (1 - 18y/u)

Substitute u = 8x^2 + 9y^2:

dy/dx = (16x/(8x^2 + 9y^2)) / (1 - 18y/(8x^2 + 9y^2))

Simplify:

dy/dx = (16x) / (8x^2 + 9y^2 - 18xy)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.