What is the derivative of y=((x^2+4)/(x^2-4))^3 ?

The derivative of y=((x^2+4)/(x^2-4))^3 is -(48x(x^2+4)^2)/((x^2-4)^4)

What is the first derivative of y=((x^2+4)/(x^2-4))^3 ?

The first derivative of y=((x^2+4)/(x^2-4))^3 is -(48x(x^2+4)^2)/((x^2-4)^4)

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derivative of

y = (\frac{x ^{2} + 4}{x ^{2} - 4})^{3}

- \frac{48 x ( x ^{2} + 4 ) ^{2}}{( x ^{2} - 4 ) ^{4}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{2} + 4}{x ^{2} - 4}) = - \frac{16 x}{( x ^{2} - 4 ) ^{2}}

= 3 (\frac{x ^{2} + 4}{x ^{2} - 4})^{2} (- \frac{16 x}{( x ^{2} - 4 ) ^{2}})

Simplify

3 (\frac{x ^{2} + 4}{x ^{2} - 4})^{2} (- \frac{16 x}{( x ^{2} - 4 ) ^{2}}) : - \frac{48 x ( x ^{2} + 4 ) ^{2}}{( x ^{2} - 4 ) ^{4}}

= - \frac{48 x ( x ^{2} + 4 ) ^{2}}{( x ^{2} - 4 ) ^{4}}

x \to 6 - lim (\frac{x}{x - 6})

\int \frac{1}{( x + 2 ) ( x + 2 )} d x

f (x) = sin^{5} (x)

\frac{d}{d t} (arcsin (2 t))

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

What is the derivative of y=((x^2+4)/(x^2-4))^3 ?
The derivative of y=((x^2+4)/(x^2-4))^3 is -(48x(x^2+4)^2)/((x^2-4)^4)
What is the first derivative of y=((x^2+4)/(x^2-4))^3 ?
The first derivative of y=((x^2+4)/(x^2-4))^3 is -(48x(x^2+4)^2)/((x^2-4)^4)