What is the derivative of (sqrt(25-x^2)/(16x)) ?

The derivative of (sqrt(25-x^2)/(16x)) is -(25)/(16x^2sqrt(25-x^2))

What is the first derivative of (sqrt(25-x^2)/(16x)) ?

The first derivative of (sqrt(25-x^2)/(16x)) is -(25)/(16x^2sqrt(25-x^2))

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\frac{d}{d x} (\frac{25 - x ^{2}}{16 x})

- \frac{25}{16 x ^{2} 25 - x ^{2}}

Solution steps

\frac{d}{d x} (\frac{25 - x ^{2}}{16 x})

Take the constant out:

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

= \frac{1}{16} \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}{16} \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}{16} \cdot \frac{( - \frac{x}{25 - x ^{2}} ) x - 1 \cdot 25 - x ^{2}}{x ^{2}}

Simplify

\frac{1}{16} \cdot \frac{( - \frac{x}{25 - x ^{2}} ) x - 1 \cdot 25 - x ^{2}}{x ^{2}} : - \frac{25}{16 x ^{2} 25 - x ^{2}}

= - \frac{25}{16 x ^{2} 25 - x ^{2}}

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

\int_{2}^{11} (x - 3)^{- \frac{2}{3}} d x

f (x) = arcsin (3 x)

∣ sin (x) ∣

n = 4 \sum \infty (\frac{1}{4})^{n}

What is the derivative of (sqrt(25-x^2)/(16x)) ?
The derivative of (sqrt(25-x^2)/(16x)) is -(25)/(16x^2sqrt(25-x^2))
What is the first derivative of (sqrt(25-x^2)/(16x)) ?
The first derivative of (sqrt(25-x^2)/(16x)) is -(25)/(16x^2sqrt(25-x^2))