What is the derivative of (sqrt(9-x^2))/(4x) ?

The derivative of (sqrt(9-x^2))/(4x) is -9/(4x^2sqrt(9-x^2))

What is the first derivative of (sqrt(9-x^2))/(4x) ?

The first derivative of (sqrt(9-x^2))/(4x) is -9/(4x^2sqrt(9-x^2))

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derivative of

\frac{9 - x ^{2}}{4 x}

- \frac{9}{4 x ^{2} 9 - x ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 - x ^{2}}{4 x})

Take the constant out:

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

= \frac{1}{4} \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}{4} \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}{4} \cdot \frac{( - \frac{x}{9 - x ^{2}} ) x - 1 \cdot 9 - x ^{2}}{x ^{2}}

Simplify

\frac{1}{4} \cdot \frac{( - \frac{x}{9 - x ^{2}} ) x - 1 \cdot 9 - x ^{2}}{x ^{2}} : - \frac{9}{4 x ^{2} 9 - x ^{2}}

= - \frac{9}{4 x ^{2} 9 - x ^{2}}

f (x) = (x + 1) (2 x - 1)

\int_{0}^{\frac{π}{8}} sin^{5} (4 x) d x

y^{^{'''}} + y^{^{''}} + 3 y^{^{'}} - 5 y = 0

\int \frac{sin ( 2 x )}{2 sin ( x )} d x

\int_{4}^{\infty} x e^{- 3 x} d x

What is the derivative of (sqrt(9-x^2))/(4x) ?
The derivative of (sqrt(9-x^2))/(4x) is -9/(4x^2sqrt(9-x^2))
What is the first derivative of (sqrt(9-x^2))/(4x) ?
The first derivative of (sqrt(9-x^2))/(4x) is -9/(4x^2sqrt(9-x^2))