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

The derivative of f(x)=((x^2+3)/(x^2-3))^4 is -(48x(x^2+3)^3)/((x^2-3)^5)

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

The first derivative of f(x)=((x^2+3)/(x^2-3))^4 is -(48x(x^2+3)^3)/((x^2-3)^5)

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derivative of

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

- \frac{48 x ( x ^{2} + 3 ) ^{3}}{( x ^{2} - 3 ) ^{5}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} - 3}) = - \frac{12 x}{( x ^{2} - 3 ) ^{2}}

= 4 (\frac{x ^{2} + 3}{x ^{2} - 3})^{3} (- \frac{12 x}{( x ^{2} - 3 ) ^{2}})

Simplify

4 (\frac{x ^{2} + 3}{x ^{2} - 3})^{3} (- \frac{12 x}{( x ^{2} - 3 ) ^{2}}) : - \frac{48 x ( x ^{2} + 3 ) ^{3}}{( x ^{2} - 3 ) ^{5}}

= - \frac{48 x ( x ^{2} + 3 ) ^{3}}{( x ^{2} - 3 ) ^{5}}

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n + 5}

\int \frac{cos ( 3 x ) cos ( 2 x )}{3} d x

\int_{0}^{\frac{π}{6}} 20 sec^{4} (x) d x

\int \frac{8 x + 6}{2 x ^{2} + 3 x - 2} d x

\int tan (3 x) d x

What is the derivative of f(x)=((x^2+3)/(x^2-3))^4 ?
The derivative of f(x)=((x^2+3)/(x^2-3))^4 is -(48x(x^2+3)^3)/((x^2-3)^5)
What is the first derivative of f(x)=((x^2+3)/(x^2-3))^4 ?
The first derivative of f(x)=((x^2+3)/(x^2-3))^4 is -(48x(x^2+3)^3)/((x^2-3)^5)