What is the derivative of y=(6+xf(x))/(sqrt(x)) ?

The derivative of y=(6+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-6}{2xsqrt(x)}

What is the first derivative of y=(6+xf(x))/(sqrt(x)) ?

The first derivative of y=(6+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-6}{2xsqrt(x)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = \frac{6 + x f ( x )}{x}

\frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 6}{2 x x}

Solution steps

\frac{d}{d x} (\frac{6 + x f ( x )}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 6 + x f ( x ) ) x - \frac{d}{d x} ( x ) ( 6 + x f ( x ) )}{( x ) ^{2}}

\frac{d}{d x} (6 + x f (x)) = x \frac{df}{d x} + f (x)

\frac{d}{d x} (x) = \frac{1}{2 x}

= \frac{( x \frac{df}{d x} + f ( x ) ) x - \frac{1}{2 x} ( 6 + x f ( x ) )}{( x ) ^{2}}

Simplify

\frac{( x \frac{df}{d x} + f ( x ) ) x - \frac{1}{2 x} ( 6 + x f ( x ) )}{( x ) ^{2}} : \frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 6}{2 x x}

= \frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 6}{2 x x}

\int y^{2} 2 y^{3} + 1 d y

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

\frac{( x ^{2} )}{2}

x \to 0 lim (\frac{1}{2 x ^{3}})

y = \frac{4 x ^{2} + 6 x}{3 x ^{3} + 2}

What is the derivative of y=(6+xf(x))/(sqrt(x)) ?
The derivative of y=(6+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-6}{2xsqrt(x)}
What is the first derivative of y=(6+xf(x))/(sqrt(x)) ?
The first derivative of y=(6+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-6}{2xsqrt(x)}