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

By signing up, you agree to our Terms of Service and Privacy Policy

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)

By signing up, you agree to our Terms of Service and Privacy Policy

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7