What is the derivative of log_{10}(x^3+9) ?

The derivative of log_{10}(x^3+9) is (3x^2)/(ln(10)(x^3+9))

What is the first derivative of log_{10}(x^3+9) ?

The first derivative of log_{10}(x^3+9) is (3x^2)/(ln(10)(x^3+9))

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\frac{d}{d x} (lo g_{10} (x^{3} + 9))

\frac{3 x ^{2}}{ln ( 10 ) ( x ^{3} + 9 )}

Solution steps

\frac{d}{d x} (lo g_{10} (x^{3} + 9))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{3} + 9 )}{ln ( 10 )})

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (x^{3} + 9))

Apply the chain rule:

\frac{1}{x ^{3} + 9} \frac{d}{d x} (x^{3} + 9)

= \frac{1}{x ^{3} + 9} \frac{d}{d x} (x^{3} + 9)

\frac{d}{d x} (x^{3} + 9) = 3 x^{2}

= \frac{1}{ln ( 10 )} \cdot \frac{1}{x ^{3} + 9} \cdot 3 x^{2}

Simplify

\frac{1}{ln ( 10 )} \cdot \frac{1}{x ^{3} + 9} \cdot 3 x^{2} : \frac{3 x ^{2}}{ln ( 10 ) ( x ^{3} + 9 )}

= \frac{3 x ^{2}}{ln ( 10 ) ( x ^{3} + 9 )}

2, [0, \frac{π}{3}]

\int \frac{1}{( 1 - x ) x} d x

x \to 0 lim (x^{s i n (3 x)})

Y = (3 - X^{2}) (X^{3} - X + 1), (2, 1)

x \to + π lim (\frac{x}{4} - \frac{π}{2})

What is the derivative of log_{10}(x^3+9) ?
The derivative of log_{10}(x^3+9) is (3x^2)/(ln(10)(x^3+9))
What is the first derivative of log_{10}(x^3+9) ?
The first derivative of log_{10}(x^3+9) is (3x^2)/(ln(10)(x^3+9))