What is the derivative of (5(1-4ln(x)))/(2x^5) ?

The derivative of (5(1-4ln(x)))/(2x^5) is -(5(-20ln(x)+9))/(2x^6)

What is the first derivative of (5(1-4ln(x)))/(2x^5) ?

The first derivative of (5(1-4ln(x)))/(2x^5) is -(5(-20ln(x)+9))/(2x^6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{5 ( 1 - 4 ln ( x ) )}{2 x ^{5}}

- \frac{5 ( - 20 ln ( x ) + 9 )}{2 x ^{6}}

Solution steps

\frac{d}{d x} (\frac{5 ( 1 - 4 ln ( x ) )}{2 x ^{5}})

Take the constant out:

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

= \frac{5}{2} \frac{d}{d x} (\frac{1 - 4 ln ( x )}{x ^{5}})

Apply the Quotient Rule:

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

= \frac{5}{2} \cdot \frac{\frac{d}{d x} ( 1 - 4 ln ( x ) ) x ^{5} - \frac{d}{d x} ( x ^{5} ) ( 1 - 4 ln ( x ) )}{( x ^{5} ) ^{2}}

\frac{d}{d x} (1 - 4 ln (x)) = - \frac{4}{x}

\frac{d}{d x} (x^{5}) = 5 x^{4}

= \frac{5}{2} \cdot \frac{( - \frac{4}{x} ) x ^{5} - 5 x ^{4} ( 1 - 4 ln ( x ) )}{( x ^{5} ) ^{2}}

Simplify

\frac{5}{2} \cdot \frac{( - \frac{4}{x} ) x ^{5} - 5 x ^{4} ( 1 - 4 ln ( x ) )}{( x ^{5} ) ^{2}} : - \frac{5 ( - 20 ln ( x ) + 9 )}{2 x ^{6}}

= - \frac{5 ( - 20 ln ( x ) + 9 )}{2 x ^{6}}

\int sin^{4} (5 θ) cos^{5} (5 θ) d θ

f (x) = sin (x)

\frac{d y}{d x}, (x - 1)^{2} + (y + 4)^{2} = 25

y^{^{'}} - y = 2 x e^{2 x}, y (0) = 1

f (4)

What is the derivative of (5(1-4ln(x)))/(2x^5) ?
The derivative of (5(1-4ln(x)))/(2x^5) is -(5(-20ln(x)+9))/(2x^6)
What is the first derivative of (5(1-4ln(x)))/(2x^5) ?
The first derivative of (5(1-4ln(x)))/(2x^5) is -(5(-20ln(x)+9))/(2x^6)