What is the derivative of (sqrt((x^2-4^3))/6) ?

The derivative of (sqrt((x^2-4^3))/6) is (x(x^2-4))/(2sqrt(x^2-4))

What is the first derivative of (sqrt((x^2-4^3))/6) ?

The first derivative of (sqrt((x^2-4^3))/6) is (x(x^2-4))/(2sqrt(x^2-4))

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Popular >

\frac{d}{d x} \frac{( x ^{2} - 4 ) ^{3}}{6}

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

Solution steps

\frac{d}{d x} \frac{( x ^{2} - 4 ) ^{3}}{6}

Take the constant out:

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

= \frac{1}{6} \frac{d}{d x} ((x^{2} - 4)^{3})

Apply the chain rule:

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

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

\frac{d}{d x} ((x^{2} - 4)^{3}) = 6 x (x^{2} - 4)^{2}

= \frac{1}{6} \cdot \frac{1}{2 ( x ^{2} - 4 ) ^{3}} \cdot 6 x (x^{2} - 4)^{2}

Simplify

\frac{1}{6} \cdot \frac{1}{2 ( x ^{2} - 4 ) ^{3}} \cdot 6 x (x^{2} - 4)^{2} : \frac{x ( x ^{2} - 4 )}{2 x ^{2} - 4}

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

\int \frac{1}{y ( y + 1 )} d y

y^{^{'}} = 2 t y (1 - y)

x, 0, 0.085

\frac{d y}{d x} + 2 y = e^{3 x}

\frac{d}{d x} (5 sin (3 x))

What is the derivative of (sqrt((x^2-4^3))/6) ?
The derivative of (sqrt((x^2-4^3))/6) is (x(x^2-4))/(2sqrt(x^2-4))
What is the first derivative of (sqrt((x^2-4^3))/6) ?
The first derivative of (sqrt((x^2-4^3))/6) is (x(x^2-4))/(2sqrt(x^2-4))