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

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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

(x + 1)^{'} = 1

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

Simplify

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

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

x \to \infty lim (\frac{3 x + 5}{6 x - 8})

\frac{d}{d x} (sin^{\frac{2}{3}} (x))

\frac{d}{d x} (4 sin (x) + 4 x cos (x))

n = 0 \sum \infty (\frac{4}{7})^{n}

\frac{d y}{d x} = \frac{( ( 1 + y ) ^{2} )}{( 1 + x ) ^{3}}, y (1) = 0

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