Question

find dy/dx when y = (3+x²)½​

Answer 1

Explanation:

y=(3+x^2)^1/2

logy=1/2 log(3+x^2)

1/y dy/dx =1/2×(2x/3+x^2)

dy/dx=y(1/2×(2x/3+x^2) )

dy/dx=(3+x^2)^1/2 (1/2×(2x/3+x^2) )

Answer 2

Answer:

dy/dx = x/((3+x²)½)

Explanation:

logy = 1/2 log ((3+x²)

1/y* dy/dx = 1/2 * (1/(3+x²) )* 2x

1/y* dy/dx = x/(3+x²)

dy/dx = y * x/(3+x²)

dy/dx = (3+x²)½ * x/(3+x²)

dy/dx = (3+x²)½ * x * (3+x²) ^-1

dy/dx = x/((3+x²)½)