What is the derivative of f(x)=(ln(x))/(x^4) ?

The derivative of f(x)=(ln(x))/(x^4) is (1-4ln(x))/(x^5)

What is the first derivative of f(x)=(ln(x))/(x^4) ?

The first derivative of f(x)=(ln(x))/(x^4) is (1-4ln(x))/(x^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = \frac{ln ( x )}{x ^{4}}

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

Solution steps

\frac{d}{d x} (\frac{ln ( x )}{x ^{4}})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{4}) = 4 x^{3}

= \frac{\frac{1}{x} x ^{4} - 4 x ^{3} ln ( x )}{( x ^{4} ) ^{2}}

Simplify

\frac{\frac{1}{x} x ^{4} - 4 x ^{3} ln ( x )}{( x ^{4} ) ^{2}} : \frac{1 - 4 ln ( x )}{x ^{5}}

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

x \to 1 - lim (2)

\frac{\partial}{\partial y} (ln (x - 9 y))

y^{^{'}} + 3 y = 9 - e^{t}

y = ∣ x - 9 ∣, y = \frac{x}{2}

x^{2} cos (x)

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