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

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

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

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

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derivative of

\frac{2 x ^{4}}{b ^{2} - x ^{2}}

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int \frac{( x ^{2} - x + 24 )}{x ^{3} + 6 x} d x

y = (5 t)^{t}

x \to 0 lim (\frac{2 sin ( x ) - sin ( 2 x )}{sin ( x ) - x cos ( x )})

\int_{0}^{4} 3 t^{4} + 3 t + 2 d t

n = 0 \sum \infty (\frac{x}{3})^{n}

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