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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derivative of

\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}}

\int π x d x

\frac{d}{d x} (i n^{2} (x + 3))

\int_{0}^{4} 2 x + 1 d x

x \to 3 + lim (\frac{6}{( x - 3 ) ^{5}})

y = \frac{4}{x ^{4}}

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))