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

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

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

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

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derivative of

f (x) = (\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} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

\int x^{2} + 6 x^{- \frac{3}{2}} d x

\frac{d}{d x} (\frac{2 - 5 x}{3 x ^{\frac{1}{3}}})

x \to \infty lim (\frac{( 25 x + 7 ) 5 x ^{2} + 2}{5 x ^{2} + 2})

\int 1 4^{x} d x

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