double derivative of ln(x^5) — step-by-step solution

1) Let's perform the differentiation:

\frac{d}{d x} (\ln (x^{5}))

We differentiate using:

\frac{d}{d x} (\ln (f)) = \frac{d}{d x} (f) \cdot \frac{1}{f}

(where f =

x^{5}

)

ⓘ The derivative of

\ln (x)

\frac{1}{x}

\frac{d}{d x} (\ln (x^{5}))

\frac{d}{d x} (x^{5}) \cdot \frac{1}{x^{5}}

\frac{d}{d x} (x^{5}) \cdot \frac{1}{x^{5}}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{d}{d x} (x^{5}) \cdot \frac{1}{x^{5}}

\frac{\frac{d}{d x} (x^{5}) \cdot 1}{x^{5}}

\frac{\frac{d}{d x} (x^{5})}{x^{5}}

■Differentiating

x^{n}

gives

n x^{n - 1}

\frac{d}{d x} (x^{5})

5 x^{4}

\frac{5 x^{4}}{x^{5}}

ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

\frac{5 x^{4}}{x^{5}}

\frac{5}{x}

\frac{5}{x}

To simplify differentiation, we rewrite

\frac{5}{x}

5 x^{-1}

2) Let's differentiate:

\frac{d}{d x} (5 x^{-1})

Apply the differentiation rule:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{n - 1}

with f =

x

ⓘ A constant factor stays in front of the derivative: that of

k u

k \cdot u'

\frac{d}{d x} (5 x^{-1})

5 \cdot \frac{d}{d x} (x^{-1})

■The derivative of

x^{n}

n x^{n - 1}

\frac{d}{d x} (x^{-1})

-1 \cdot x^{-2}

5 \cdot (-1 \cdot x^{-2})

ⓘ A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

5 \cdot (-1)

-5

-5 x^{-2}

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

- 5 x^{-2}

- \frac{5}{x^{2}}

▶Derivative:

- \frac{5}{x^{2}}