What is the derivative of (sqrt(9-x^2)/(16x)) ?

The derivative of (sqrt(9-x^2)/(16x)) is -9/(16x^2sqrt(9-x^2))

What is the first derivative of (sqrt(9-x^2)/(16x)) ?

The first derivative of (sqrt(9-x^2)/(16x)) is -9/(16x^2sqrt(9-x^2))

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\frac{d}{d x} (\frac{9 - x ^{2}}{16 x})

- \frac{9}{16 x ^{2} 9 - x ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 - x ^{2}}{16 x})

Take the constant out:

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

= \frac{1}{16} \frac{d}{d x} (\frac{9 - 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} ( 9 - x ^{2} ) x - \frac{d x}{d x} 9 - x ^{2}}{x ^{2}}

\frac{d}{d x} (9 - x^{2}) = - \frac{x}{9 - x ^{2}}

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

= \frac{1}{16} \cdot \frac{( - \frac{x}{9 - x ^{2}} ) x - 1 \cdot 9 - x ^{2}}{x ^{2}}

Simplify

\frac{1}{16} \cdot \frac{( - \frac{x}{9 - x ^{2}} ) x - 1 \cdot 9 - x ^{2}}{x ^{2}} : - \frac{9}{16 x ^{2} 9 - x ^{2}}

= - \frac{9}{16 x ^{2} 9 - x ^{2}}

\frac{\partial}{\partial x} (7 x sin (4 x^{2} y))

\int \frac{( z ^{2} + 1 ) ^{2}}{z} d z

f (x) = x^{3} - 3 x + 1, (2, 3)

\int cos^{3} (5 - x) sin (5 - x) d x

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

What is the derivative of (sqrt(9-x^2)/(16x)) ?
The derivative of (sqrt(9-x^2)/(16x)) is -9/(16x^2sqrt(9-x^2))
What is the first derivative of (sqrt(9-x^2)/(16x)) ?
The first derivative of (sqrt(9-x^2)/(16x)) is -9/(16x^2sqrt(9-x^2))