What is the derivative of log_{8}(2x^3+3x^2+1) ?

The derivative of log_{8}(2x^3+3x^2+1) is (2x(x+1))/(ln(2)(2x^3+3x^2+1))

What is the first derivative of log_{8}(2x^3+3x^2+1) ?

The first derivative of log_{8}(2x^3+3x^2+1) is (2x(x+1))/(ln(2)(2x^3+3x^2+1))

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derivative of

lo g_{8} (2 x^{3} + 3 x^{2} + 1)

\frac{2 x ( x + 1 )}{ln ( 2 ) ( 2 x ^{3} + 3 x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (lo g_{8} (2 x^{3} + 3 x^{2} + 1))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( 2 x ^{3} + 3 x ^{2} + 1 )}{ln ( 8 )})

Take the constant out:

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

= \frac{1}{ln ( 8 )} \frac{d}{d x} (ln (2 x^{3} + 3 x^{2} + 1))

Apply the chain rule:

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

= \frac{1}{2 x ^{3} + 3 x ^{2} + 1} \frac{d}{d x} (2 x^{3} + 3 x^{2} + 1)

\frac{d}{d x} (2 x^{3} + 3 x^{2} + 1) = 6 x^{2} + 6 x

= \frac{1}{ln ( 8 )} \cdot \frac{1}{2 x ^{3} + 3 x ^{2} + 1} (6 x^{2} + 6 x)

Simplify

\frac{1}{ln ( 8 )} \cdot \frac{1}{2 x ^{3} + 3 x ^{2} + 1} (6 x^{2} + 6 x) : \frac{2 x ( x + 1 )}{ln ( 2 ) ( 2 x ^{3} + 3 x ^{2} + 1 )}

= \frac{2 x ( x + 1 )}{ln ( 2 ) ( 2 x ^{3} + 3 x ^{2} + 1 )}

\int \frac{4 x ^{2} - 9}{x ^{3}} d x

x \to 0 lim (x)

x \to 1 lim (\frac{1}{∣ x - 1 ∣})

\frac{\partial}{\partial y} (tan (3 + 2 x^{2} y^{3} z^{2}))

\int \frac{1}{x ^{2} x - 1} d x

What is the derivative of log_{8}(2x^3+3x^2+1) ?
The derivative of log_{8}(2x^3+3x^2+1) is (2x(x+1))/(ln(2)(2x^3+3x^2+1))
What is the first derivative of log_{8}(2x^3+3x^2+1) ?
The first derivative of log_{8}(2x^3+3x^2+1) is (2x(x+1))/(ln(2)(2x^3+3x^2+1))