Answer:
dy/dx = (1 / x^3 + x) × (3x² + 1) × (1/2)
Step-by-step 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)
