What is the ((x^2-2)/(5x^3+6))^' ?

The ((x^2-2)/(5x^3+6))^' is (-5x^4+30x^2+12x)/((5x^3+6)^2)

What is the first ((x^2-2)/(5x^3+6))^' ?

The first ((x^2-2)/(5x^3+6))^' is (-5x^4+30x^2+12x)/((5x^3+6)^2)

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(\frac{x ^{2} - 2}{5 x ^{3} + 6})^{^{'}}

\frac{- 5 x ^{4} + 30 x ^{2} + 12 x}{( 5 x ^{3} + 6 ) ^{2}}

Solution steps

(\frac{x ^{2} - 2}{5 x ^{3} + 6})^{'}

Apply the Quotient Rule:

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

= \frac{( x ^{2} - 2 ) ^{^{'}} ( 5 x ^{3} + 6 ) - ( 5 x ^{3} + 6 ) ^{^{'}} ( x ^{2} - 2 )}{( 5 x ^{3} + 6 ) ^{2}}

(x^{2} - 2)^{'} = 2 x

(5 x^{3} + 6)^{'} = 15 x^{2}

= \frac{2 x ( 5 x ^{3} + 6 ) - 15 x ^{2} ( x ^{2} - 2 )}{( 5 x ^{3} + 6 ) ^{2}}

Expand

2 x (5 x^{3} + 6) - 15 x^{2} (x^{2} - 2) : - 5 x^{4} + 30 x^{2} + 12 x

= \frac{- 5 x ^{4} + 30 x ^{2} + 12 x}{( 5 x ^{3} + 6 ) ^{2}}

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 5, y^{^{'}} (0) = 10

\frac{\partial}{\partial x} (tanh (x))

x \to 0 lim (\frac{tan ( 5 x )}{x})

\int \frac{1}{x ^{3}} - x d x

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

What is the ((x^2-2)/(5x^3+6))^' ?
The ((x^2-2)/(5x^3+6))^' is (-5x^4+30x^2+12x)/((5x^3+6)^2)
What is the first ((x^2-2)/(5x^3+6))^' ?
The first ((x^2-2)/(5x^3+6))^' is (-5x^4+30x^2+12x)/((5x^3+6)^2)