What is the derivative of f(x)=(x+7)/(8x) ?

The derivative of f(x)=(x+7)/(8x) is -7/(8x^2)

What is the first derivative of f(x)=(x+7)/(8x) ?

The first derivative of f(x)=(x+7)/(8x) is -7/(8x^2)

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derivative of

f (x) = \frac{x + 7}{8 x}

- \frac{7}{8 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x + 7}{8 x})

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (\frac{x + 7}{x})

Apply the Quotient Rule:

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

= \frac{1}{8} \cdot \frac{\frac{d}{d x} ( x + 7 ) x - \frac{d x}{d x} ( x + 7 )}{x ^{2}}

\frac{d}{d x} (x + 7) = 1

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

= \frac{1}{8} \cdot \frac{1 \cdot x - 1 \cdot ( x + 7 )}{x ^{2}}

Simplify

\frac{1}{8} \cdot \frac{1 \cdot x - 1 \cdot ( x + 7 )}{x ^{2}} : - \frac{7}{8 x ^{2}}

= - \frac{7}{8 x ^{2}}

\frac{d}{d x} (\frac{1}{sin ^{3} ( x )})

y = \frac{7}{x}

\frac{\partial}{\partial x} (x^{2} (x - 2)^{4})

x^{2} y^{^{''}} + 3 x y^{^{'}} - 3 y = 0, y (1) = 0, y^{^{'}} (1) = 5

\int_{0}^{\frac{π}{4}} x sin (12 x) d x

What is the derivative of f(x)=(x+7)/(8x) ?
The derivative of f(x)=(x+7)/(8x) is -7/(8x^2)
What is the first derivative of f(x)=(x+7)/(8x) ?
The first derivative of f(x)=(x+7)/(8x) is -7/(8x^2)