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

The derivative of ln(ln(2x+2y)) is (1+\frac{dy)/(dx)}{ln(2x+2y)(x+y)}

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

The first derivative of ln(ln(2x+2y)) is (1+\frac{dy)/(dx)}{ln(2x+2y)(x+y)}

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\frac{d}{d x} (ln (ln (2 x + 2 y)))

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

Solution steps

\frac{d}{d x} (ln (ln (2 x + 2 y)))

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int 0.2 x^{2} + 1 d x

n = 1 \sum \infty \frac{8}{5 ^{n - 1}}

\frac{d}{d θ} (r (θ)), r (θ) = cos (θ) cot (θ)

f (x) = (x^{3} + 2) (x^{5} + 7 x^{2} + 3)

What is the derivative of ln(ln(2x+2y)) ?
The derivative of ln(ln(2x+2y)) is (1+\frac{dy)/(dx)}{ln(2x+2y)(x+y)}
What is the first derivative of ln(ln(2x+2y)) ?
The first derivative of ln(ln(2x+2y)) is (1+\frac{dy)/(dx)}{ln(2x+2y)(x+y)}