What is the derivative of 1/(ln(x+1)-1/(ln(x+3))) ?

The derivative of 1/(ln(x+1)-1/(ln(x+3))) is -1/(ln^2(x+1)(x+1))+1/(ln^2(x+3)(x+3))

What is the first derivative of 1/(ln(x+1)-1/(ln(x+3))) ?

The first derivative of 1/(ln(x+1)-1/(ln(x+3))) is -1/(ln^2(x+1)(x+1))+1/(ln^2(x+3)(x+3))

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\frac{d}{d x} (\frac{1}{ln ( x + 1 )} - \frac{1}{ln ( x + 3 )})

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

Solution steps

\frac{d}{d x} (\frac{1}{ln ( x + 1 )} - \frac{1}{ln ( x + 3 )})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1}{ln ( x + 1 )}) - \frac{d}{d x} (\frac{1}{ln ( x + 3 )})

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

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

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

Simplify

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

x \to 0 lim (x \cdot csc (x))

x \to 0 lim (5 x - 2)

\frac{d}{d x} (ln (x^{4} + cot (3 x)))

sin^{2} (e^{s i n (2) (t)})

\frac{d}{d x} (2 cos^{4} (x))

What is the derivative of 1/(ln(x+1)-1/(ln(x+3))) ?
The derivative of 1/(ln(x+1)-1/(ln(x+3))) is -1/(ln^2(x+1)(x+1))+1/(ln^2(x+3)(x+3))
What is the first derivative of 1/(ln(x+1)-1/(ln(x+3))) ?
The first derivative of 1/(ln(x+1)-1/(ln(x+3))) is -1/(ln^2(x+1)(x+1))+1/(ln^2(x+3)(x+3))