Question

Find dy/dx............

Answer 1

dy/dx = (1 / x^3 + x) × (3x² + 1) × (1/2)

Explanation:

y = log[ x² × √(x² + 1) ]

y = log[ √(x(x² + 1)) ]

y = log[ √(x^3 + x) ]

y = log[ √(x^3 + x) ]

Now, let a = √(x^3 + x)

Then y = log(a)

Find dy/da.

y = log(a)

dy/da = (1 / a)

dy/da = (1 / √(x^3 + x))

Find da/dx using chain rule.

a = √(x^3 + x)

Let b = x^3 + x, then a = √b

da/dx = (db / dx) × (da / db)

da/dx = (3x² + 1) × (1/2)× (b)^(-1/2)

da/dx = (3x² + 1) × (1/2)× (x^3 + x)^(-1/2)

Finally, find dy/dx using chain rule.

dy/dx = (dy/da) × (da/dx)

dy/dx = (1 / √(x^3 + x)) × (3x² + 1) × (1/2)×

(x^3 + x)^(-1/2)

dy/dx = (1 / (x^3 + x)) × (3x² + 1) × (1/2)