What is the derivative of 8x^4ln(6x) ?

The derivative of 8x^4ln(6x) is 8(4x^3ln(6x)+x^3)

What is the first derivative of 8x^4ln(6x) ?

The first derivative of 8x^4ln(6x) is 8(4x^3ln(6x)+x^3)

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derivative of

8 x^{4} ln (6 x)

8 (4 x^{3} ln (6 x) + x^{3})

Solution steps

\frac{d}{d x} (8 x^{4} ln (6 x))

Take the constant out:

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

= 8 \frac{d}{d x} (x^{4} ln (6 x))

Apply the Product Rule:

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

f = x^{4}, g = ln (6 x)

= 8 (\frac{d}{d x} (x^{4}) ln (6 x) + \frac{d}{d x} (ln (6 x)) x^{4})

\frac{d}{d x} (x^{4}) = 4 x^{3}

\frac{d}{d x} (ln (6 x)) = \frac{1}{x}

= 8 (4 x^{3} ln (6 x) + \frac{1}{x} x^{4})

\frac{1}{x} x^{4} = x^{3}

= 8 (4 x^{3} ln (6 x) + x^{3})

\int cot (x) cos^{2} (x) d x

\frac{d}{d x} ((x^{2} - 1)^{4})

\int \frac{x ^{2} + 4}{( x - 5 ) ( x ^{2} + 9 )} d x

x \to 0 lim (ln (x) \cdot x)

\int \frac{x ^{2} + 1}{( x - 4 ) ( x - 3 ) ^{2}} d x

What is the derivative of 8x^4ln(6x) ?
The derivative of 8x^4ln(6x) is 8(4x^3ln(6x)+x^3)
What is the first derivative of 8x^4ln(6x) ?
The first derivative of 8x^4ln(6x) is 8(4x^3ln(6x)+x^3)