What is the derivative of y=(sqrt(25-x^2))/(4x) ?

The derivative of y=(sqrt(25-x^2))/(4x) is -(25)/(4x^2sqrt(25-x^2))

What is the first derivative of y=(sqrt(25-x^2))/(4x) ?

The first derivative of y=(sqrt(25-x^2))/(4x) is -(25)/(4x^2sqrt(25-x^2))

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derivative of

y = \frac{25 - x ^{2}}{4 x}

- \frac{25}{4 x ^{2} 25 - x ^{2}}

Solution steps

\frac{d}{d x} (\frac{25 - x ^{2}}{4 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{4} \frac{d}{d x} (\frac{25 - x ^{2}}{x})

Apply the Quotient Rule:

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

= \frac{1}{4} \cdot \frac{\frac{d}{d x} ( 25 - x ^{2} ) x - \frac{d x}{d x} 25 - x ^{2}}{x ^{2}}

\frac{d}{d x} (25 - x^{2}) = - \frac{x}{25 - x ^{2}}

\frac{d x}{d x} = 1

= \frac{1}{4} \cdot \frac{( - \frac{x}{25 - x ^{2}} ) x - 1 \cdot 25 - x ^{2}}{x ^{2}}

Simplify

\frac{1}{4} \cdot \frac{( - \frac{x}{25 - x ^{2}} ) x - 1 \cdot 25 - x ^{2}}{x ^{2}} : - \frac{25}{4 x ^{2} 25 - x ^{2}}

= - \frac{25}{4 x ^{2} 25 - x ^{2}}

x y^{2} + 3 y^{2} - x^{2} \cdot \frac{d y}{d x} = 0

y^{^{'}} - y = t e^{3 t}

\int \frac{1}{x + 1} d x

x \to - 1 lim (\frac{ln ( x )}{x - 1})

x \to - 2 - lim ((3 x^{2}) - 6 x + 2)

What is the derivative of y=(sqrt(25-x^2))/(4x) ?
The derivative of y=(sqrt(25-x^2))/(4x) is -(25)/(4x^2sqrt(25-x^2))
What is the first derivative of y=(sqrt(25-x^2))/(4x) ?
The first derivative of y=(sqrt(25-x^2))/(4x) is -(25)/(4x^2sqrt(25-x^2))