What is the derivative of (2x-1/(3x^2)) ?

The derivative of (2x-1/(3x^2)) is (2(-x+1))/(3x^3)

What is the first derivative of (2x-1/(3x^2)) ?

The first derivative of (2x-1/(3x^2)) is (2(-x+1))/(3x^3)

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\frac{d}{d x} (\frac{2 x - 1}{3 x ^{2}})

\frac{2 ( - x + 1 )}{3 x ^{3}}

Solution steps

\frac{d}{d x} (\frac{2 x - 1}{3 x ^{2}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{3} \frac{d}{d x} (\frac{2 x - 1}{x ^{2}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{3} \cdot \frac{\frac{d}{d x} ( 2 x - 1 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 2 x - 1 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (2 x - 1) = 2

\frac{d}{d x} (x^{2}) = 2 x

= \frac{1}{3} \cdot \frac{2 x ^{2} - 2 x ( 2 x - 1 )}{( x ^{2} ) ^{2}}

Simplify

\frac{1}{3} \cdot \frac{2 x ^{2} - 2 x ( 2 x - 1 )}{( x ^{2} ) ^{2}} : \frac{2 ( - x + 1 )}{3 x ^{3}}

= \frac{2 ( - x + 1 )}{3 x ^{3}}

x y^{^{'}} = y + x e^{\frac{y}{x}}

e^{3 x}, e^{x}, [0, 1]

\int_{- 6}^{8} 10 d x

\int_{- 4}^{2} - x^{2} - 2 x + 8 d x

\int \frac{2 arctan ( x )}{x ^{3}} d x

What is the derivative of (2x-1/(3x^2)) ?
The derivative of (2x-1/(3x^2)) is (2(-x+1))/(3x^3)
What is the first derivative of (2x-1/(3x^2)) ?
The first derivative of (2x-1/(3x^2)) is (2(-x+1))/(3x^3)