What is the derivative of (x^3-ax/(bx)) ?

The derivative of (x^3-ax/(bx)) is (2x)/b

What is the first derivative of (x^3-ax/(bx)) ?

The first derivative of (x^3-ax/(bx)) is (2x)/b

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\frac{d}{d x} (\frac{x ^{3} - a x}{b x})

\frac{2 x}{b}

Solution steps

\frac{d}{d x} (\frac{x ^{3} - a x}{b x})

Take the constant out:

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

= \frac{1}{b} \frac{d}{d x} (\frac{x ^{3} - a x}{x})

Apply the Quotient Rule:

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

= \frac{1}{b} \cdot \frac{\frac{d}{d x} ( x ^{3} - a x ) x - \frac{d x}{d x} ( x ^{3} - a x )}{x ^{2}}

\frac{d}{d x} (x^{3} - a x) = 3 x^{2} - a

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

= \frac{1}{b} \cdot \frac{( 3 x ^{2} - a ) x - 1 \cdot ( x ^{3} - a x )}{x ^{2}}

Simplify

\frac{1}{b} \cdot \frac{( 3 x ^{2} - a ) x - 1 \cdot ( x ^{3} - a x )}{x ^{2}} : \frac{2 x}{b}

= \frac{2 x}{b}

f (t) = sin^{3} (t)

y^{^{''}} - y^{^{'}} - 2 y = 0

f (t) = 7 sin (4 (t + 2)) + 3 cos (4 t - 2)

\int \frac{2 x}{x ^{2} + 3} d x

\int sin^{- 4} (\frac{x}{5}) cos^{3} (\frac{x}{5}) d x

What is the derivative of (x^3-ax/(bx)) ?
The derivative of (x^3-ax/(bx)) is (2x)/b
What is the first derivative of (x^3-ax/(bx)) ?
The first derivative of (x^3-ax/(bx)) is (2x)/b