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

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

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

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

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

2 (2 x^{2 x} (ln (x) + 1)^{2} + x^{2 x - 1})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2 x}, g = ln (x) + 1

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

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

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

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

Simplify

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

= 2 (2 x^{2 x} (ln (x) + 1)^{2} + x^{2 x - 1})

\frac{1}{1 + 2 x ^{2}}

\frac{d}{d x} (x^{\frac{2}{3}} \cdot (x - 4)^{2})

\frac{d}{d x} (\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}})

\int (x (sin (x^{2}))) (cos^{8} (x^{2})) d x

\frac{d}{d x} (81 cos (3 x))

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