What is the derivative of sqrt(((x^2-4^3)/(36))) ?

The derivative of sqrt(((x^2-4^3)/(36))) is (xsqrt(x^2-4))/2

What is the first derivative of sqrt(((x^2-4^3)/(36))) ?

The first derivative of sqrt(((x^2-4^3)/(36))) is (xsqrt(x^2-4))/2

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\frac{d}{d x} \frac{( x ^{2} - 4 ) ^{3}}{36}

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

Solution steps

\frac{d}{d x} \frac{( x ^{2} - 4 ) ^{3}}{36}

Apply the chain rule:

\frac{1}{2 \frac{( x ^{2} - 4 ) ^{3}}{36}} \frac{d}{d x} (\frac{( x ^{2} - 4 ) ^{3}}{36})

= \frac{1}{2 \frac{( x ^{2} - 4 ) ^{3}}{36}} \frac{d}{d x} (\frac{( x ^{2} - 4 ) ^{3}}{36})

\frac{d}{d x} (\frac{( x ^{2} - 4 ) ^{3}}{36}) = \frac{x ( x ^{2} - 4 ) ^{2}}{6}

= \frac{1}{2 \frac{( x ^{2} - 4 ) ^{3}}{36}} \cdot \frac{x ( x ^{2} - 4 ) ^{2}}{6}

Simplify

\frac{1}{2 \frac{( x ^{2} - 4 ) ^{3}}{36}} \cdot \frac{x ( x ^{2} - 4 ) ^{2}}{6} : \frac{x x ^{2} - 4}{2}

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

x \to 1 + lim (\frac{1}{x ^{3}})

(- 5, - 1), (0, 3)

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

n = 1 \sum \infty \frac{( 3 x ) ^{n}}{n}

\int x - x d x

What is the derivative of sqrt(((x^2-4^3)/(36))) ?
The derivative of sqrt(((x^2-4^3)/(36))) is (xsqrt(x^2-4))/2
What is the first derivative of sqrt(((x^2-4^3)/(36))) ?
The first derivative of sqrt(((x^2-4^3)/(36))) is (xsqrt(x^2-4))/2