What is the derivative of (5x^4/(sqrt(2-x^2))) ?

The derivative of (5x^4/(sqrt(2-x^2))) is (5(8x^3-3x^5))/((2-x^2)sqrt(2-x^2))

What is the first derivative of (5x^4/(sqrt(2-x^2))) ?

The first derivative of (5x^4/(sqrt(2-x^2))) is (5(8x^3-3x^5))/((2-x^2)sqrt(2-x^2))

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\frac{d}{d x} (\frac{5 x ^{4}}{2 - x ^{2}})

\frac{5 ( 8 x ^{3} - 3 x ^{5} )}{( 2 - x ^{2} ) 2 - x ^{2}}

Solution steps

\frac{d}{d x} (\frac{5 x ^{4}}{2 - x ^{2}})

Take the constant out:

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

= 5 \frac{d}{d x} (\frac{x ^{4}}{2 - x ^{2}})

Apply the Quotient Rule:

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

= 5 \cdot \frac{\frac{d}{d x} ( x ^{4} ) 2 - x ^{2} - \frac{d}{d x} ( 2 - x ^{2} ) x ^{4}}{( 2 - x ^{2} ) ^{2}}

\frac{d}{d x} (x^{4}) = 4 x^{3}

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

= 5 \cdot \frac{4 x ^{3} 2 - x ^{2} - ( - \frac{x}{2 - x ^{2}} ) x ^{4}}{( 2 - x ^{2} ) ^{2}}

Simplify

5 \cdot \frac{4 x ^{3} 2 - x ^{2} - ( - \frac{x}{2 - x ^{2}} ) x ^{4}}{( 2 - x ^{2} ) ^{2}} : \frac{5 ( 8 x ^{3} - 3 x ^{5} )}{( 2 - x ^{2} ) 2 - x ^{2}}

= \frac{5 ( 8 x ^{3} - 3 x ^{5} )}{( 2 - x ^{2} ) 2 - x ^{2}}

x \to \infty lim (\frac{x ^{2} - 9}{2 x - 6})

y^{^{'}} - y tan (x) = cos (x)

\frac{\partial}{\partial s} (sin (s + t))

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x}

\frac{2}{5} x^{2} + \frac{1}{4} x - 1

What is the derivative of (5x^4/(sqrt(2-x^2))) ?
The derivative of (5x^4/(sqrt(2-x^2))) is (5(8x^3-3x^5))/((2-x^2)sqrt(2-x^2))
What is the first derivative of (5x^4/(sqrt(2-x^2))) ?
The first derivative of (5x^4/(sqrt(2-x^2))) is (5(8x^3-3x^5))/((2-x^2)sqrt(2-x^2))