What is the derivative of f(x)= 6/(ln(x^2+4)) ?

The derivative of f(x)= 6/(ln(x^2+4)) is -(12x)/(ln^2(x^2+4)(x^2+4))

What is the first derivative of f(x)= 6/(ln(x^2+4)) ?

The first derivative of f(x)= 6/(ln(x^2+4)) is -(12x)/(ln^2(x^2+4)(x^2+4))

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derivative of

f (x) = \frac{6}{ln ( x ^{2} + 4 )}

- \frac{12 x}{ln ^{2} ( x ^{2} + 4 ) ( x ^{2} + 4 )}

Solution steps

\frac{d}{d x} (\frac{6}{ln ( x ^{2} + 4 )})

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 6 \frac{d}{d x} (ln^{- 1} (x^{2} + 4))

Apply the chain rule:

- \frac{1}{( ln ( x ^{2} + 4 ) ) ^{2}} \frac{d}{d x} (ln (x^{2} + 4))

= - \frac{1}{( ln ( x ^{2} + 4 ) ) ^{2}} \frac{d}{d x} (ln (x^{2} + 4))

\frac{d}{d x} (ln (x^{2} + 4)) = \frac{2 x}{x ^{2} + 4}

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

Simplify

6 (- \frac{1}{ln ^{2} ( x ^{2} + 4 )} \cdot \frac{2 x}{x ^{2} + 4}) : - \frac{12 x}{ln ^{2} ( x ^{2} + 4 ) ( x ^{2} + 4 )}

= - \frac{12 x}{ln ^{2} ( x ^{2} + 4 ) ( x ^{2} + 4 )}

\int \frac{sin ( 44 x )}{1 + cos ^{2} ( 44 x )} d x

\int 2 x (x^{2} + 1)^{- 7} d x

f (x) = x^{2}, [0, 3]

\int_{- 1}^{3} (x^{2} + 6 x + 3) d x

\frac{d ^{3}}{d x ^{3}} (x^{4} - 10 x^{3})

What is the derivative of f(x)= 6/(ln(x^2+4)) ?
The derivative of f(x)= 6/(ln(x^2+4)) is -(12x)/(ln^2(x^2+4)(x^2+4))
What is the first derivative of f(x)= 6/(ln(x^2+4)) ?
The first derivative of f(x)= 6/(ln(x^2+4)) is -(12x)/(ln^2(x^2+4)(x^2+4))