What is the derivative of log_{7}(3x^2+1) ?

The derivative of log_{7}(3x^2+1) is (6x)/(ln(7)(3x^2+1))

What is the first derivative of log_{7}(3x^2+1) ?

The first derivative of log_{7}(3x^2+1) is (6x)/(ln(7)(3x^2+1))

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\frac{d}{d x} (lo g_{7} (3 x^{2} + 1))

\frac{6 x}{ln ( 7 ) ( 3 x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (lo g_{7} (3 x^{2} + 1))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 3 x ^{2} + 1 )}{ln ( 7 )})

Take the constant out:

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

= \frac{1}{ln ( 7 )} \frac{d}{d x} (ln (3 x^{2} + 1))

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{2} + 1) = 6 x

= \frac{1}{ln ( 7 )} \cdot \frac{1}{3 x ^{2} + 1} \cdot 6 x

Simplify

\frac{1}{ln ( 7 )} \cdot \frac{1}{3 x ^{2} + 1} \cdot 6 x : \frac{6 x}{ln ( 7 ) ( 3 x ^{2} + 1 )}

= \frac{6 x}{ln ( 7 ) ( 3 x ^{2} + 1 )}

5 + x^{2}

\int e^{- s t} sin (3 t) d t

t \to 0 lim (sin^{t} (t))

y = sin ((π x)^{4})

\frac{\partial}{\partial x} (x^{3} - 7 x^{2} + y^{3} - 5 y)

What is the derivative of log_{7}(3x^2+1) ?
The derivative of log_{7}(3x^2+1) is (6x)/(ln(7)(3x^2+1))
What is the first derivative of log_{7}(3x^2+1) ?
The first derivative of log_{7}(3x^2+1) is (6x)/(ln(7)(3x^2+1))