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

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

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

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

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derivative of

y = (\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}}

x \to 1 lim (- x)

\frac{( 1 + x ) d y}{d x} - x y = x + x^{2}

\frac{e ^{7 x}}{ln ( 6 x )}

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

\frac{d}{d x} (\frac{1}{6 + x ^{2}})

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