What is the derivative of 1/(((1+x)^2ln^2(x+1))) ?

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

5 x^{6}

2 e^{3} t sin (2 t)

6 x^{4} - 7 x^{2} + 9

3 x + y^{2}

tan (e^{s}) in x^{2}

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