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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{x}{1 + x ^{2}}, at x = - 1

y = 2 - x, y = 4 - x^{2}

\int (x^{2} + e^{1 - x}) d x

y^{^{''}} + 2 y + y = x e^{- x}

\int - \frac{1}{3} e^{3 x} d x

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