What is the derivative of y=ln(1/(xsqrt(x+1))) ?

The derivative of y=ln(1/(xsqrt(x+1))) is -(3x+2)/(2x(x+1))

What is the first derivative of y=ln(1/(xsqrt(x+1))) ?

The first derivative of y=ln(1/(xsqrt(x+1))) is -(3x+2)/(2x(x+1))

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derivative of

y = ln (\frac{1}{x x + 1})

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

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply the chain rule:

\frac{1}{x x + 1} \frac{d}{d x} (x x + 1)

= \frac{1}{x x + 1} \frac{d}{d x} (x x + 1)

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

= - \frac{1}{x x + 1} \cdot \frac{3 x + 2}{2 x + 1}

Simplify

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

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

x y^{^{'}} - 2 y - 16 = 0

y^{^{'}} = e^{5 y} sin (6 x)

x y^{^{'}} - 10 y = 0

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

x \to 0 lim (\frac{- 6 + x}{x ^{4}})

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