What is the derivative of ((3x^2+1^3)/((2x^2-1)^4)) ?

The derivative of ((3x^2+1^3)/((2x^2-1)^4)) is (2x(3x^2+1)^2(-6x^2-17))/((2x^2-1)^5)

What is the first derivative of ((3x^2+1^3)/((2x^2-1)^4)) ?

The first derivative of ((3x^2+1^3)/((2x^2-1)^4)) is (2x(3x^2+1)^2(-6x^2-17))/((2x^2-1)^5)

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\frac{d}{d x} (\frac{( 3 x ^{2} + 1 ) ^{3}}{( 2 x ^{2} - 1 ) ^{4}})

\frac{2 x ( 3 x ^{2} + 1 ) ^{2} ( - 6 x ^{2} - 17 )}{( 2 x ^{2} - 1 ) ^{5}}

Solution steps

\frac{d}{d x} (\frac{( 3 x ^{2} + 1 ) ^{3}}{( 2 x ^{2} - 1 ) ^{4}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ( 3 x ^{2} + 1 ) ^{3} ) ( 2 x ^{2} - 1 ) ^{4} - \frac{d}{d x} ( ( 2 x ^{2} - 1 ) ^{4} ) ( 3 x ^{2} + 1 ) ^{3}}{( ( 2 x ^{2} - 1 ) ^{4} ) ^{2}}

\frac{d}{d x} ((3 x^{2} + 1)^{3}) = 18 x (3 x^{2} + 1)^{2}

\frac{d}{d x} ((2 x^{2} - 1)^{4}) = 16 x (2 x^{2} - 1)^{3}

= \frac{18 x ( 3 x ^{2} + 1 ) ^{2} ( 2 x ^{2} - 1 ) ^{4} - 16 x ( 2 x ^{2} - 1 ) ^{3} ( 3 x ^{2} + 1 ) ^{3}}{( ( 2 x ^{2} - 1 ) ^{4} ) ^{2}}

Simplify

\frac{18 x ( 3 x ^{2} + 1 ) ^{2} ( 2 x ^{2} - 1 ) ^{4} - 16 x ( 2 x ^{2} - 1 ) ^{3} ( 3 x ^{2} + 1 ) ^{3}}{( ( 2 x ^{2} - 1 ) ^{4} ) ^{2}} : \frac{2 x ( 3 x ^{2} + 1 ) ^{2} ( - 6 x ^{2} - 17 )}{( 2 x ^{2} - 1 ) ^{5}}

= \frac{2 x ( 3 x ^{2} + 1 ) ^{2} ( - 6 x ^{2} - 17 )}{( 2 x ^{2} - 1 ) ^{5}}

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

n = 1 \sum \infty (\frac{n - 2}{n})^{n}

\int (sin (2 u) - cos (3 u))^{(2)} d u

\frac{d}{d x} (- 3 x - 9)

\int \frac{y}{( 1 + y ^{2} ) ^{4}} d y

What is the derivative of ((3x^2+1^3)/((2x^2-1)^4)) ?
The derivative of ((3x^2+1^3)/((2x^2-1)^4)) is (2x(3x^2+1)^2(-6x^2-17))/((2x^2-1)^5)
What is the first derivative of ((3x^2+1^3)/((2x^2-1)^4)) ?
The first derivative of ((3x^2+1^3)/((2x^2-1)^4)) is (2x(3x^2+1)^2(-6x^2-17))/((2x^2-1)^5)