What is the derivative of y=log_{3}(x^2+1) ?

The derivative of y=log_{3}(x^2+1) is (2x)/(ln(3)(x^2+1))

What is the first derivative of y=log_{3}(x^2+1) ?

The first derivative of y=log_{3}(x^2+1) is (2x)/(ln(3)(x^2+1))

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derivative of

y = lo g_{3} (x^{2} + 1)

\frac{2 x}{ln ( 3 ) ( x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (lo g_{3} (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 ( 3 )})

Take the constant out:

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

= \frac{1}{ln ( 3 )} \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 ( 3 )} \cdot \frac{1}{x ^{2} + 1} \cdot 2 x

Simplify

\frac{1}{ln ( 3 )} \cdot \frac{1}{x ^{2} + 1} \cdot 2 x : \frac{2 x}{ln ( 3 ) ( x ^{2} + 1 )}

= \frac{2 x}{ln ( 3 ) ( x ^{2} + 1 )}

\frac{d y}{d x} = \frac{26}{x}

f (x) = x^{4} + 9 x

\int cos (x) sin (y) d x

x \to \infty lim (4, x)

\frac{\partial}{\partial x} (sec^{2} (2 x + 3 y) \cdot 3)

What is the derivative of y=log_{3}(x^2+1) ?
The derivative of y=log_{3}(x^2+1) is (2x)/(ln(3)(x^2+1))
What is the first derivative of y=log_{3}(x^2+1) ?
The first derivative of y=log_{3}(x^2+1) is (2x)/(ln(3)(x^2+1))