What is the derivative of (8x^2-x)ln(x) ?

The derivative of (8x^2-x)ln(x) is ln(x)(16x-1)+8x-1

What is the first derivative of (8x^2-x)ln(x) ?

The first derivative of (8x^2-x)ln(x) is ln(x)(16x-1)+8x-1

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derivative of

(8 x^{2} - x) ln (x)

ln (x) (16 x - 1) + 8 x - 1

Solution steps

\frac{d}{d x} ((8 x^{2} - x) ln (x))

Apply the Product Rule:

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

f = 8 x^{2} - x, g = ln (x)

= \frac{d}{d x} (8 x^{2} - x) ln (x) + \frac{d}{d x} (ln (x)) (8 x^{2} - x)

\frac{d}{d x} (8 x^{2} - x) = 16 x - 1

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

= (16 x - 1) ln (x) + \frac{1}{x} (8 x^{2} - x)

\frac{1}{x} (8 x^{2} - x) = 8 x - 1

= ln (x) (16 x - 1) + 8 x - 1

What is the derivative of (8x^2-x)ln(x) ?
The derivative of (8x^2-x)ln(x) is ln(x)(16x-1)+8x-1
What is the first derivative of (8x^2-x)ln(x) ?
The first derivative of (8x^2-x)ln(x) is ln(x)(16x-1)+8x-1