What is the derivative of log_{10}(x^2+x+1) ?

The derivative of log_{10}(x^2+x+1) is (2x+1)/(ln(10)(x^2+x+1))

What is the first derivative of log_{10}(x^2+x+1) ?

The first derivative of log_{10}(x^2+x+1) is (2x+1)/(ln(10)(x^2+x+1))

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\frac{d}{d x} (lo g_{10} (x^{2} + x + 1))

\frac{2 x + 1}{ln ( 10 ) ( x ^{2} + x + 1 )}

Solution steps

\frac{d}{d x} (lo g_{10} (x^{2} + x + 1))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{2} + x + 1 )}{ln ( 10 )})

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (x^{2} + x + 1))

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = \frac{2 x + 3}{3 x - 2}

\int_{0}^{2} \frac{1}{8 + 2 x ^{2}} d x

3 ln (x), x ln (x), 1, 2

\frac{d y}{d x} = - \frac{e ^{x}}{8}

\int sec^{2} (3 x) tan^{3} (3 x) d x

What is the derivative of log_{10}(x^2+x+1) ?
The derivative of log_{10}(x^2+x+1) is (2x+1)/(ln(10)(x^2+x+1))
What is the first derivative of log_{10}(x^2+x+1) ?
The first derivative of log_{10}(x^2+x+1) is (2x+1)/(ln(10)(x^2+x+1))