What is the derivative of f(x)=ln((sqrt(4+x^2))/x) ?

The derivative of f(x)=ln((sqrt(4+x^2))/x) is -4/(x(x^2+4))

What is the first derivative of f(x)=ln((sqrt(4+x^2))/x) ?

The first derivative of f(x)=ln((sqrt(4+x^2))/x) is -4/(x(x^2+4))

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derivative of

f (x) = ln (\frac{4 + x ^{2}}{x})

- \frac{4}{x ( x ^{2} + 4 )}

Solution steps

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

Apply the chain rule:

\frac{1}{\frac{4 + x ^{2}}{x}} \frac{d}{d x} (\frac{4 + x ^{2}}{x})

= \frac{1}{\frac{4 + x ^{2}}{x}} \frac{d}{d x} (\frac{4 + x ^{2}}{x})

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

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

Simplify

\frac{1}{\frac{4 + x ^{2}}{x}} (- \frac{4}{x ^{2} 4 + x ^{2}}) : - \frac{4}{x ( x ^{2} + 4 )}

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

f (x) = 20 x, (4, 40)

y^{^{''}} + 5 y^{^{'}} - 2 y = 0

\int \frac{1}{t ^{2} - 12 t + 40} d t

x \to 0 - lim \frac{1}{\frac{1}{x}}

(t^{6} + 26 t^{4} - t^{3} + 179)^{73}

What is the derivative of f(x)=ln((sqrt(4+x^2))/x) ?
The derivative of f(x)=ln((sqrt(4+x^2))/x) is -4/(x(x^2+4))
What is the first derivative of f(x)=ln((sqrt(4+x^2))/x) ?
The first derivative of f(x)=ln((sqrt(4+x^2))/x) is -4/(x(x^2+4))