What is the derivative of y=ln(1/(xsqrt(x+4))) ?

The derivative of y=ln(1/(xsqrt(x+4))) is -(3x+8)/(2x(x+4))

What is the first derivative of y=ln(1/(xsqrt(x+4))) ?

The first derivative of y=ln(1/(xsqrt(x+4))) is -(3x+8)/(2x(x+4))

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derivative of

y = ln (\frac{1}{x x + 4})

- \frac{3 x + 8}{2 x ( x + 4 )}

Solution steps

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

Simplify

ln (\frac{1}{x x + 4}) : - ln (x x + 4)

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

Take the constant out:

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

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

Apply the chain rule:

\frac{1}{x x + 4} \frac{d}{d x} (x x + 4)

= \frac{1}{x x + 4} \frac{d}{d x} (x x + 4)

\frac{d}{d x} (x x + 4) = \frac{3 x + 8}{2 x + 4}

= - \frac{1}{x x + 4} \cdot \frac{3 x + 8}{2 x + 4}

Simplify

- \frac{1}{x x + 4} \cdot \frac{3 x + 8}{2 x + 4} : - \frac{3 x + 8}{2 x ( x + 4 )}

= - \frac{3 x + 8}{2 x ( x + 4 )}

f (x) = x^{3}

(- 43, - 4)

x^{2} e^{x}

e^{- 2 x}

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

What is the derivative of y=ln(1/(xsqrt(x+4))) ?
The derivative of y=ln(1/(xsqrt(x+4))) is -(3x+8)/(2x(x+4))
What is the first derivative of y=ln(1/(xsqrt(x+4))) ?
The first derivative of y=ln(1/(xsqrt(x+4))) is -(3x+8)/(2x(x+4))