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How to differentiate with respect to X?

0 votes

 

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

asked Mar 14, 2013 in CALCULUS by Jose Rodriguez Rookie
reshown Mar 14, 2013 by moderator

1 Answer

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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(xn-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

 

answered Mar 15, 2013 by diane Scholar

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