What is the derivative of sqrt(1-(x^2/(y^2))) ?

The derivative of sqrt(1-(x^2/(y^2))) is -(x(-x\frac{dy)/(dx)+y)}{y^2sqrt(-x^2+y^2)}

What is the first derivative of sqrt(1-(x^2/(y^2))) ?

The first derivative of sqrt(1-(x^2/(y^2))) is -(x(-x\frac{dy)/(dx)+y)}{y^2sqrt(-x^2+y^2)}

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (1 - \frac{x ^{2}}{y ^{2}}) = - \frac{2 x ( - x \frac{d y}{d x} + y )}{y ^{3}}

= \frac{1}{2 1 - \frac{x ^{2}}{y ^{2}}} - \frac{2 x ( - x \frac{d y}{d x} + y )}{y ^{3}}

Simplify

\frac{1}{2 1 - \frac{x ^{2}}{y ^{2}}} - \frac{2 x ( - x \frac{d y}{d x} + y )}{y ^{3}} : - \frac{x ( - x \frac{d y}{d x} + y )}{y ^{2} - x ^{2} + y ^{2}}

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

(1 + x)^{k}, 0

\int 1.830496064 e^{2.1342 x} sin (x) d x

\int 4 cos (2 x) d x

\frac{\partial}{\partial y} (x y^{2} + 1)

y^{^{''}} - 43 y^{^{'}} + 12 y = 0

What is the derivative of sqrt(1-(x^2/(y^2))) ?
The derivative of sqrt(1-(x^2/(y^2))) is -(x(-x\frac{dy)/(dx)+y)}{y^2sqrt(-x^2+y^2)}
What is the first derivative of sqrt(1-(x^2/(y^2))) ?
The first derivative of sqrt(1-(x^2/(y^2))) is -(x(-x\frac{dy)/(dx)+y)}{y^2sqrt(-x^2+y^2)}