What is the derivative of y=((x^2+3)/(x^2-3))^8 ?

The derivative of y=((x^2+3)/(x^2-3))^8 is -(96x(x^2+3)^7)/((x^2-3)^9)

What is the first derivative of y=((x^2+3)/(x^2-3))^8 ?

The first derivative of y=((x^2+3)/(x^2-3))^8 is -(96x(x^2+3)^7)/((x^2-3)^9)

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derivative of

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

- \frac{96 x ( x ^{2} + 3 ) ^{7}}{( x ^{2} - 3 ) ^{9}}

Solution steps

\frac{d}{d x} ((\frac{x ^{2} + 3}{x ^{2} - 3})^{8})

Apply the chain rule:

8 (\frac{x ^{2} + 3}{x ^{2} - 3})^{7} \frac{d}{d x} (\frac{x ^{2} + 3}{x ^{2} - 3})

= 8 (\frac{x ^{2} + 3}{x ^{2} - 3})^{7} \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}}

= 8 (\frac{x ^{2} + 3}{x ^{2} - 3})^{7} (- \frac{12 x}{( x ^{2} - 3 ) ^{2}})

Simplify

8 (\frac{x ^{2} + 3}{x ^{2} - 3})^{7} (- \frac{12 x}{( x ^{2} - 3 ) ^{2}}) : - \frac{96 x ( x ^{2} + 3 ) ^{7}}{( x ^{2} - 3 ) ^{9}}

= - \frac{96 x ( x ^{2} + 3 ) ^{7}}{( x ^{2} - 3 ) ^{9}}

f (x) = \frac{7}{1 - x}

x^{2} x^{3} + 1

x \to 8 lim (x)

\int y^{3} (4 y^{2} - 25) d y

n = 4 \sum \infty \frac{1}{n}

What is the derivative of y=((x^2+3)/(x^2-3))^8 ?
The derivative of y=((x^2+3)/(x^2-3))^8 is -(96x(x^2+3)^7)/((x^2-3)^9)
What is the first derivative of y=((x^2+3)/(x^2-3))^8 ?
The first derivative of y=((x^2+3)/(x^2-3))^8 is -(96x(x^2+3)^7)/((x^2-3)^9)