What is the derivative of 1/((3x^2+x-3^9)) ?

The derivative of 1/((3x^2+x-3^9)) is -(9(6x+1))/((3x^2+x-3)^{10)}

What is the first derivative of 1/((3x^2+x-3^9)) ?

The first derivative of 1/((3x^2+x-3^9)) is -(9(6x+1))/((3x^2+x-3)^{10)}

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\frac{d}{d x} (\frac{1}{( 3 x ^{2} + x - 3 ) ^{9}})

- \frac{9 ( 6 x + 1 )}{( 3 x ^{2} + x - 3 ) ^{10}}

Solution steps

\frac{d}{d x} (\frac{1}{( 3 x ^{2} + x - 3 ) ^{9}})

Apply exponent rule:

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

= \frac{d}{d x} ((3 x^{2} + x - 3)^{- 9})

Apply the chain rule:

- \frac{9}{( 3 x ^{2} + x - 3 ) ^{10}} \frac{d}{d x} (3 x^{2} + x - 3)

= - \frac{9}{( 3 x ^{2} + x - 3 ) ^{10}} \frac{d}{d x} (3 x^{2} + x - 3)

\frac{d}{d x} (3 x^{2} + x - 3) = 6 x + 1

= - \frac{9}{( 3 x ^{2} + x - 3 ) ^{10}} (6 x + 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{9 ( 6 x + 1 )}{( 3 x ^{2} + x - 3 ) ^{10}}

f (x) = x^{x^{2}}

x \to 9 lim (25 - x)

\frac{\partial}{\partial x} (2 x y e^{x^{2}})

x \to 1 lim ((x - 2 [x])^{2})

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- x} - 5 x

What is the derivative of 1/((3x^2+x-3^9)) ?
The derivative of 1/((3x^2+x-3^9)) is -(9(6x+1))/((3x^2+x-3)^{10)}
What is the first derivative of 1/((3x^2+x-3^9)) ?
The first derivative of 1/((3x^2+x-3^9)) is -(9(6x+1))/((3x^2+x-3)^{10)}