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

The derivative of 2/((x^2-3x^2)) is -(4(2x-3))/((x^2-3x)^3)

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

The first derivative of 2/((x^2-3x^2)) is -(4(2x-3))/((x^2-3x)^3)

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

- \frac{4 ( 2 x - 3 )}{( x ^{2} - 3 x ) ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

2 (- \frac{2}{( x ^{2} - 3 x ) ^{3}} (2 x - 3)) : - \frac{4 ( 2 x - 3 )}{( x ^{2} - 3 x ) ^{3}}

= - \frac{4 ( 2 x - 3 )}{( x ^{2} - 3 x ) ^{3}}

\frac{d}{d x} ((x^{2} + a^{2})^{5})

\frac{\partial}{\partial x} (2 x - 5 y + 3)

\frac{d}{d x} (6 e^{- 4 x} ln (x))

n = 1 \sum \infty (\frac{7}{6})^{n}

x y - 2 y + x y^{^{'}} - y^{^{'}} = 0

What is the derivative of 2/((x^2-3x^2)) ?
The derivative of 2/((x^2-3x^2)) is -(4(2x-3))/((x^2-3x)^3)
What is the first derivative of 2/((x^2-3x^2)) ?
The first derivative of 2/((x^2-3x^2)) is -(4(2x-3))/((x^2-3x)^3)