What is the derivative of ln(ln(x^2+1)) ?

The derivative of ln(ln(x^2+1)) is (2x)/(ln(x^2+1)(x^2+1))

What is the first derivative of ln(ln(x^2+1)) ?

The first derivative of ln(ln(x^2+1)) is (2x)/(ln(x^2+1)(x^2+1))

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

\frac{2 x}{ln ( x ^{2} + 1 ) ( x ^{2} + 1 )}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{''}} + a^{2} y = 0

x y^{^{''}} + 3 y^{^{'}} = 0

\frac{\partial}{\partial x} (ln (x^{2} - y))

\frac{d}{d y} (\frac{y ^{3}}{8})

t an g e n t y = x^{4} + 8 x^{2} - x, (1, 8)

What is the derivative of ln(ln(x^2+1)) ?
The derivative of ln(ln(x^2+1)) is (2x)/(ln(x^2+1)(x^2+1))
What is the first derivative of ln(ln(x^2+1)) ?
The first derivative of ln(ln(x^2+1)) is (2x)/(ln(x^2+1)(x^2+1))