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

The derivative of (y^2+x^2)/(y^2-x^2) is -(4x^2y)/((y^2-x^2)^2)

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

The first derivative of (y^2+x^2)/(y^2-x^2) is -(4x^2y)/((y^2-x^2)^2)

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derivative of

\frac{y ^{2} + x ^{2}}{y ^{2} - x ^{2}}

- \frac{4 x ^{2} y}{( y ^{2} - x ^{2} ) ^{2}}

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{2 y ( y ^{2} - x ^{2} ) - 2 y ( y ^{2} + x ^{2} )}{( y ^{2} - x ^{2} ) ^{2}} : - \frac{4 x ^{2} y}{( y ^{2} - x ^{2} ) ^{2}}

= - \frac{4 x ^{2} y}{( y ^{2} - x ^{2} ) ^{2}}

f (x) = \frac{2}{x}, at x = 1

x^{2} - 3 x + 3

\int 2 y e^{2 z} d z

h (g^{^{'}} (x))

y = \frac{4}{x}, y = 16 x, y = \frac{1}{16} x

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