What is the derivative of f(x)=log_{4}(3x^2+1) ?

The derivative of f(x)=log_{4}(3x^2+1) is (3x)/(ln(2)(3x^2+1))

What is the first derivative of f(x)=log_{4}(3x^2+1) ?

The first derivative of f(x)=log_{4}(3x^2+1) is (3x)/(ln(2)(3x^2+1))

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derivative of

f (x) = lo g_{4} (3 x^{2} + 1)

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

Solution steps

\frac{d}{d x} (lo g_{4} (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 ( 4 )})

Take the constant out:

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

= \frac{1}{ln ( 4 )} \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 ( 4 )} \cdot \frac{1}{3 x ^{2} + 1} \cdot 6 x

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{3 x ^{2} + 1} \cdot 6 x : \frac{3 x}{ln ( 2 ) ( 3 x ^{2} + 1 )}

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

x \to 14 - lim (15 x + ln (14 - x))

(1 + e^{- t})^{2}

y = x^{s i n (x)}

t^{2} - 8 t + 16

\frac{d}{d x} ((2 x)^{6 x})

What is the derivative of f(x)=log_{4}(3x^2+1) ?
The derivative of f(x)=log_{4}(3x^2+1) is (3x)/(ln(2)(3x^2+1))
What is the first derivative of f(x)=log_{4}(3x^2+1) ?
The first derivative of f(x)=log_{4}(3x^2+1) is (3x)/(ln(2)(3x^2+1))