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

The derivative of f(x)=ln((x^2)/(x^2+4)) is 8/(x(x^2+4))

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

The first derivative of f(x)=ln((x^2)/(x^2+4)) is 8/(x(x^2+4))

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derivative of

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

\frac{8}{x ( x ^{2} + 4 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} + 4}) = \frac{8 x}{( x ^{2} + 4 ) ^{2}}

= \frac{1}{\frac{x ^{2}}{x ^{2} + 4}} \cdot \frac{8 x}{( x ^{2} + 4 ) ^{2}}

Simplify

\frac{1}{\frac{x ^{2}}{x ^{2} + 4}} \cdot \frac{8 x}{( x ^{2} + 4 ) ^{2}} : \frac{8}{x ( x ^{2} + 4 )}

= \frac{8}{x ( x ^{2} + 4 )}

\int \frac{1}{( 15 - x ) ( 60 - x )} d x

x \to - 1 lim (\frac{k x + 3}{4})

y = x^{3} - 3 x^{2} + 4

\frac{d}{d x} (2 (6 - x^{2})^{5})

\frac{\partial}{\partial x} (sin (x) \cdot cos (y))

What is the derivative of f(x)=ln((x^2)/(x^2+4)) ?
The derivative of f(x)=ln((x^2)/(x^2+4)) is 8/(x(x^2+4))
What is the first derivative of f(x)=ln((x^2)/(x^2+4)) ?
The first derivative of f(x)=ln((x^2)/(x^2+4)) is 8/(x(x^2+4))