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

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

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

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

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derivative of

y = lo g_{10} (x^{2} + 2 x + 1)

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

Solution steps

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

Apply log rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{1}{3 tan ^{3} ( x )}

(x + 2)^{- 4}

\frac{d}{d y} (2 y)

\frac{1}{( 3 x + 4 ) ^{3}}

1 - 1 - x^{2}

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