What is the derivative of (2x/(sqrt(1+4x^2+9x^4))) ?

The derivative of (2x/(sqrt(1+4x^2+9x^4))) is (2(-9x^4+1))/((1+4x^2+9x^4)sqrt(1+4x^2+9x^4))

What is the first derivative of (2x/(sqrt(1+4x^2+9x^4))) ?

The first derivative of (2x/(sqrt(1+4x^2+9x^4))) is (2(-9x^4+1))/((1+4x^2+9x^4)sqrt(1+4x^2+9x^4))

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\frac{d}{d x} (\frac{2 x}{1 + 4 x ^{2} + 9 x ^{4}})

\frac{2 ( - 9 x ^{4} + 1 )}{( 1 + 4 x ^{2} + 9 x ^{4} ) 1 + 4 x ^{2} + 9 x ^{4}}

Solution steps

\frac{d}{d x} (\frac{2 x}{1 + 4 x ^{2} + 9 x ^{4}})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{x}{1 + 4 x ^{2} + 9 x ^{4}})

Apply the Quotient Rule:

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

= 2 \cdot \frac{\frac{d x}{d x} 1 + 4 x ^{2} + 9 x ^{4} - \frac{d}{d x} ( 1 + 4 x ^{2} + 9 x ^{4} ) x}{( 1 + 4 x ^{2} + 9 x ^{4} ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (1 + 4 x^{2} + 9 x^{4}) = \frac{2 x ( 9 x ^{2} + 2 )}{1 + 4 x ^{2} + 9 x ^{4}}

= 2 \cdot \frac{1 \cdot 1 + 4 x ^{2} + 9 x ^{4} - \frac{2 x ( 9 x ^{2} + 2 )}{1 + 4 x ^{2} + 9 x ^{4}} x}{( 1 + 4 x ^{2} + 9 x ^{4} ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot 1 + 4 x ^{2} + 9 x ^{4} - \frac{2 x ( 9 x ^{2} + 2 )}{1 + 4 x ^{2} + 9 x ^{4}} x}{( 1 + 4 x ^{2} + 9 x ^{4} ) ^{2}} : \frac{2 ( - 9 x ^{4} + 1 )}{( 1 + 4 x ^{2} + 9 x ^{4} ) 1 + 4 x ^{2} + 9 x ^{4}}

= \frac{2 ( - 9 x ^{4} + 1 )}{( 1 + 4 x ^{2} + 9 x ^{4} ) 1 + 4 x ^{2} + 9 x ^{4}}

\frac{d}{d x} (cos^{2} (x^{4}))

\int_{0}^{2} \frac{x}{8} d x

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

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

\int \frac{x ^{2}}{x ^{2} + 1} d x

What is the derivative of (2x/(sqrt(1+4x^2+9x^4))) ?
The derivative of (2x/(sqrt(1+4x^2+9x^4))) is (2(-9x^4+1))/((1+4x^2+9x^4)sqrt(1+4x^2+9x^4))
What is the first derivative of (2x/(sqrt(1+4x^2+9x^4))) ?
The first derivative of (2x/(sqrt(1+4x^2+9x^4))) is (2(-9x^4+1))/((1+4x^2+9x^4)sqrt(1+4x^2+9x^4))