If y = log(x2√(x2 + 1)) then find dy/dx.

1.	If y = log(x2√(x2 + 1)) then find dy/dx.
Answer» y = log(x²√(x²+ 1)) Differentiating w.r.t. 'x' dy/dx = ((d log(x²√(x² + 1)))/d(x²√(x² + 2))) x ((d(x²√(x² + 1)))/dx) = (1/x²√(x² + 1)) x ((dx²/dx) x √(x² + 1) + (x²d√(x² + 1)/dx)) = (1/x²√(x² + 1)) x (2x√(x² + 1) + x² x (1/2√(x² + 1)) x 2x) = (1/x²√(x² + 1)) x (2x√(x² + 1) + (x²/√(x² + 1)))

Answer»

y = log(x²√(x²+ 1))

Differentiating w.r.t. 'x'

dy/dx = ((d log(x²√(x² + 1)))/d(x²√(x² + 2))) x ((d(x²√(x² + 1)))/dx)

= (1/x²√(x² + 1)) x ((dx²/dx) x √(x² + 1) + (x²d√(x² + 1)/dx))

= (1/x²√(x² + 1)) x (2x√(x² + 1) + x² x (1/2√(x² + 1)) x 2x)

= (1/x²√(x² + 1)) x (2x√(x² + 1) + (x²/√(x² + 1)))

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