What is the (log_{10}(1+x))^' ?

The (log_{10}(1+x))^' is 1/(ln(10)(1+x))

What is the first (log_{10}(1+x))^' ?

The first (log_{10}(1+x))^' is 1/(ln(10)(1+x))

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(lo g_{10} (1 + x))^{^{'}}

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

Solution steps

(lo g_{10} (1 + x))^{'}

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

\frac{1}{1 + x} (1 + x)^{'}

= \frac{1}{1 + x} (1 + x)^{'}

(1 + x)^{'} = 1

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

Simplify

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

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

- 8 x

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