What is the derivative of y=log_{2}(log_{2}(x)) ?

The derivative of y=log_{2}(log_{2}(x)) is 1/(ln^2(2)xlog_{2)(x)}

What is the first derivative of y=log_{2}(log_{2}(x)) ?

The first derivative of y=log_{2}(log_{2}(x)) is 1/(ln^2(2)xlog_{2)(x)}

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derivative of

y = lo g_{2} (lo g_{2} (x))

\frac{1}{ln ^{2} ( 2 ) x lo g _{2} ( x )}

Solution steps

\frac{d}{d x} (lo g_{2} (lo g_{2} (x)))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( lo g _{2} ( x ) )}{ln ( 2 )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{ln ( 2 )} \frac{d}{d x} (ln (lo g_{2} (x)))

Apply the chain rule:

\frac{1}{lo g _{2} ( x )} \frac{d}{d x} (lo g_{2} (x))

= \frac{1}{lo g _{2} ( x )} \frac{d}{d x} (lo g_{2} (x))

\frac{d}{d x} (lo g_{2} (x)) = \frac{1}{x ln ( 2 )}

= \frac{1}{ln ( 2 )} \cdot \frac{1}{lo g _{2} ( x )} \cdot \frac{1}{x ln ( 2 )}

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{lo g _{2} ( x )} \cdot \frac{1}{x ln ( 2 )} : \frac{1}{ln ^{2} ( 2 ) x lo g _{2} ( x )}

= \frac{1}{ln ^{2} ( 2 ) x lo g _{2} ( x )}

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

(1 - x y) d x + (x y - x^{2}) d y = 0

\int tan^{2} (x) x d x

\int_{3}^{11} x d x

n = 0 \sum \infty n!

What is the derivative of y=log_{2}(log_{2}(x)) ?
The derivative of y=log_{2}(log_{2}(x)) is 1/(ln^2(2)xlog_{2)(x)}
What is the first derivative of y=log_{2}(log_{2}(x)) ?
The first derivative of y=log_{2}(log_{2}(x)) is 1/(ln^2(2)xlog_{2)(x)}