What is the derivative of f(x)=(ln(x))/(20+x) ?

The derivative of f(x)=(ln(x))/(20+x) is (20+x-xln(x))/(x(20+x)^2)

What is the first derivative of f(x)=(ln(x))/(20+x) ?

The first derivative of f(x)=(ln(x))/(20+x) is (20+x-xln(x))/(x(20+x)^2)

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derivative of

f (x) = \frac{ln ( x )}{20 + x}

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

Solution steps

\frac{d}{d x} (\frac{ln ( x )}{20 + x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ln ( x ) ) ( 20 + x ) - \frac{d}{d x} ( 20 + x ) ln ( x )}{( 20 + x ) ^{2}}

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

\frac{d}{d x} (20 + x) = 1

= \frac{\frac{1}{x} ( 20 + x ) - 1 \cdot ln ( x )}{( 20 + x ) ^{2}}

Simplify

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

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

\frac{d ^{2}}{d x ^{2}} (z^{2} e^{z})

y^{^{'}} - 2 y = e^{t}

x \to 4 lim (x^{2} - 10)

x \to 1 lim (\frac{6}{x ^{3} - 1})

\int (- y) d y

What is the derivative of f(x)=(ln(x))/(20+x) ?
The derivative of f(x)=(ln(x))/(20+x) is (20+x-xln(x))/(x(20+x)^2)
What is the first derivative of f(x)=(ln(x))/(20+x) ?
The first derivative of f(x)=(ln(x))/(20+x) is (20+x-xln(x))/(x(20+x)^2)