What is the derivative of xlog_{e}(2x) ?

The derivative of xlog_{e}(2x) is ln(2x)+1

What is the first derivative of xlog_{e}(2x) ?

The first derivative of xlog_{e}(2x) is ln(2x)+1

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

ln (2 x) + 1

Solution steps

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

Simplify

x lo g_{e} (2 x) : x ln (2 x)

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

Apply the Product Rule:

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

f = x, g = ln (2 x)

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

\frac{d x}{d x} = 1

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

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

Simplify

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

= ln (2 x) + 1

y^{''} (t)

f (x) = 3 - 2 x e^{x}, at x = 0

f (x) = (10 x - 4)^{\frac{1}{2}}, at x = 2

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

f (y) = 6 x^{2} ln (x^{2})

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