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

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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

n = 0 \sum \infty n (- 1)^{n}

\frac{\partial}{\partial u} (u \cdot v)

y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y = 1.73 sin (33 π x)

\frac{d}{d x} (cos (k x + b))

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