How to differentiate with respect to X?

Question

 

2x^3 + the square root of x +(x^2 +2x) over x^2

Answer 1

2x³+ √(x) + (x² + 2x) / x²

Put y = 2x³+ √(x) + (x² + 2x) / x²

Diferentiate each side with respective x

dy/dx = d/dx [2x³+√(x) + (x³+ 2x) / x²]

The derivative formulas

d/dx (xⁿ) = n(x^n-1) and d /dx[√(x)] = 1 / 2√(x)

dy/dx = d/dx [2x³] + d/dx[√(x)] + d /dx[(x² + 2x) / x²]

[x² + 2x) / x²] = x² / x² + 2x / x²

[x² + 2x) / x²] = 1+ 2 / x

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

d / dx(1+ 2 / x) = 0 + 2(-1 / x²) (Using derivative formula)

dy/dx = 6x² + 1 / 2√(x) + (0 + 2(-1 / x²)

dy/dx = 6x^2 + 1 / 2√(x) +(-2 / x^2)

dy/dx = 6x^2 + 1 / 2√(x) - 2 / x²