What is the derivative of (9x^2+9sqrt(x))/(8x) ?

The derivative of (9x^2+9sqrt(x))/(8x) is (18x^2-9sqrt(x))/(16x^2)

What is the first derivative of (9x^2+9sqrt(x))/(8x) ?

The first derivative of (9x^2+9sqrt(x))/(8x) is (18x^2-9sqrt(x))/(16x^2)

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derivative of

\frac{9 x ^{2} + 9 x}{8 x}

\frac{18 x ^{2} - 9 x}{16 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 x ^{2} + 9 x}{8 x})

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (\frac{9 x ^{2} + 9 x}{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} ( 9 x ^{2} + 9 x ) x - \frac{d x}{d x} ( 9 x ^{2} + 9 x )}{x ^{2}}

\frac{d}{d x} (9 x^{2} + 9 x) = 18 x + \frac{9}{2 x}

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

= \frac{1}{8} \cdot \frac{( 18 x + \frac{9}{2 x} ) x - 1 \cdot ( 9 x ^{2} + 9 x )}{x ^{2}}

Simplify

\frac{1}{8} \cdot \frac{( 18 x + \frac{9}{2 x} ) x - 1 \cdot ( 9 x ^{2} + 9 x )}{x ^{2}} : \frac{18 x ^{2} - 9 x}{16 x ^{2}}

= \frac{18 x ^{2} - 9 x}{16 x ^{2}}

\int \frac{1}{2 - t} + \frac{1}{2 + t} d t

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

y^{^{''}} + y^{^{'}} = sin^{2} (x)

x = - 2, x = 1, y = 11 x, y = x^{2} - 12

\frac{d}{d x} (400 - 5 x)

What is the derivative of (9x^2+9sqrt(x))/(8x) ?
The derivative of (9x^2+9sqrt(x))/(8x) is (18x^2-9sqrt(x))/(16x^2)
What is the first derivative of (9x^2+9sqrt(x))/(8x) ?
The first derivative of (9x^2+9sqrt(x))/(8x) is (18x^2-9sqrt(x))/(16x^2)