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

The derivative of ((x^2+4)/(x^2-4))^4 is -(64x(x^2+4)^3)/((x^2-4)^5)

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

The first derivative of ((x^2+4)/(x^2-4))^4 is -(64x(x^2+4)^3)/((x^2-4)^5)

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derivative of

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

- \frac{64 x ( x ^{2} + 4 ) ^{3}}{( x ^{2} - 4 ) ^{5}}

Solution steps

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

Apply the chain rule:

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

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

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

Simplify

4 (\frac{x ^{2} + 4}{x ^{2} - 4})^{3} (- \frac{16 x}{( x ^{2} - 4 ) ^{2}}) : - \frac{64 x ( x ^{2} + 4 ) ^{3}}{( x ^{2} - 4 ) ^{5}}

= - \frac{64 x ( x ^{2} + 4 ) ^{3}}{( x ^{2} - 4 ) ^{5}}

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

y = \frac{4 x + 5}{x - 1}, (2, 13)

\frac{d}{d x} (a sec (x))

\int \frac{ln ( x ^{6} )}{x} d x

\int e^{- s t} cos (b t) d t

What is the derivative of ((x^2+4)/(x^2-4))^4 ?
The derivative of ((x^2+4)/(x^2-4))^4 is -(64x(x^2+4)^3)/((x^2-4)^5)
What is the first derivative of ((x^2+4)/(x^2-4))^4 ?
The first derivative of ((x^2+4)/(x^2-4))^4 is -(64x(x^2+4)^3)/((x^2-4)^5)