The equation of a curve is xy^3-2x^2y^2=0. find the gradient of the tangent to the curve at (1,2)

Differentiate both sides of

xy ³ - 2x ²y ² = 0

with respect to x :

d(xy ³ - 2x ²y ²)/dx = d(0)/dx

d(xy ³)/dx - 2 d(x ²y ²)/dx = 0

By the product rule,

d(x)/dx y ³ + x d(y ³)/dx - 2 (d(x ²)/dx y ² + x ² d(y ²)/dx) = 0

By the chain rule,

y ³ + 3xy ² dy/dx - 2 (2xy ² + 2x ²y dy/dx) = 0

y ³ + 3xy ² dy/dx - 4xy ² - 4x ²y dy/dx = 0

y ³ - 4xy ² + (3xy ² - 4x ²y) dy/dx = 0

(3xy ² - 4x ²y) dy/dx = 4xy ² - y ³

dy/dx = (4xy ² - y ³) / (3xy ² - 4x ²y)

dy/dx = (4xy - y ²) / (3xy - 4x ²)

At the point (1, 2), the gradient is

dy/dx (1, 2) = (4×1×2 - 2²) / (3×1×2 - 4×1²) = 4/2 = 2

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm