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

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

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

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

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derivative of

y = (\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}}

n = 1 \sum \infty \frac{1}{n ^{2}}

\int_{1}^{2} \frac{5}{x ^{3} + 3 x} d x

\int ((3 x^{5} - 4)^{2} \cdot 5 x^{4}) d x

\frac{\partial}{\partial x} (tan (x yz))

\frac{\partial}{\partial x} (x^{7} y^{6} - x^{6} y^{5})

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