derivative of f(x)=x^x(ln(x)+1) — step-by-step solution

We save the definition:

f (x) = x^{x} \cdot (\ln (x) + 1)

1) Let's perform the differentiation:

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

We differentiate

u1

x^{x}

and

v2

\ln (x) + 1

using:

\frac{d}{d x} (u1 \cdot v2) = \frac{d}{d x} (u1) \cdot v2 + u1 \cdot \frac{d}{d x} (v2)

2) Let's differentiate

u1

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

To differentiate on intervals where

x > 0

, write

x^{x}

e^{x \cdot \ln (x)}

ⅰ

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

\frac{d}{d x} (x \cdot \ln (x)) \cdot e^{x \cdot \ln (x)}

\frac{d}{d x} (x \cdot \ln (x)) \cdot e^{x \cdot \ln (x)}

⇥For

u3

x

and

v4

\ln (x)

, use the differentiation rule:

\frac{d}{d x} (u3 \cdot v4) = \frac{d}{d x} (u3) \cdot v4 + u3 \cdot \frac{d}{d x} (v4)

⇥2.1) Let's differentiate

u3

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

⇥ⓘ The derivative of x is 1:

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

1

⇥

1

⇥2.2) Let's differentiate

v4

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

⇥ⓘ The derivative of ln(x) is 1/x:

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

\frac{1}{x}

⇥

\frac{1}{x}

⇥2.3) We evaluate the resulting derivative (

u3 \cdot v4

)':

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

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

x \cdot \frac{1}{x}

\frac{x \cdot 1}{x}

\frac{x}{x}

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

x

\frac{x}{x}

1

⇥

\ln (x) + 1

⇥2.4) Then you can resume the derivation:

ⅰ

\frac{d}{d x} (x \cdot \ln (x))

\ln (x) + 1

(\ln (x) + 1) \cdot x^{x}

ⓘ Expand by multiplying each term inside the parentheses:

(\ln (x) + 1) \cdot x^{x}

\ln (x) \cdot x^{x} + x^{x}

\ln (x) \cdot x^{x} + x^{x}

3) Let's perform the differentiation

v2

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

\frac{d}{d x} (\ln (x)) + 0

\frac{d}{d x} (\ln (x)) + 0

ⓘ Adding 0 changes nothing:

\frac{d}{d x} (\ln (x)) + 0

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

■The derivative of ln(x) is 1/x:

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

\frac{1}{x}

\frac{1}{x}

4) Let's evaluate the resulting derivative (

u1 \cdot v2

)':

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

ⓘ Multiplying a fraction by a term multiplies its numerator:

x^{x} \cdot \frac{1}{x}

\frac{x^{x} \cdot 1}{x}

\frac{x^{x}}{x}

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

\frac{x^{x}}{x}

x^{x - 1}

(\ln (x) \cdot x^{x} + x^{x}) \cdot (\ln (x) + 1) + x^{x - 1}

ⓘ Expand by multiplying each term inside the parentheses:

(\ln (x) \cdot x^{x} + x^{x}) \cdot (\ln (x) + 1)

\ln (x) \cdot x^{x} \cdot \ln (x)

+ \ln (x) \cdot x^{x} \cdot 1

+ x^{x} \cdot \ln (x)

+ x^{x} \cdot 1

\ln (x) \cdot x^{x} \cdot \ln (x)

+ \ln (x) \cdot x^{x} \cdot 1

+ x^{x} \cdot \ln (x)

+ x^{x} \cdot 1

+ x^{x - 1}

ⓘ Multiplying powers with the same base adds the exponents:

\ln (x) \cdot \ln (x)

{\ln (x)}^{2}

{\ln (x)}^{2} \cdot x^{x}

+ \ln (x) \cdot x^{x}

+ x^{x} \cdot \ln (x)

+ x^{x}

+ x^{x - 1}

ⓘ Like terms are grouped by adding their coefficients:

\ln (x) \cdot x^{x} + x^{x} \cdot \ln (x)

2 \cdot \ln (x) \cdot x^{x}

{\ln (x)}^{2} \cdot x^{x}

+ 2 \cdot \ln (x) \cdot x^{x}

+ x^{x}

+ x^{x - 1}

5) You can then resume the derivation:

ⅰ

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

{\ln (x)}^{2} \cdot x^{x}

+ 2 \cdot \ln (x) \cdot x^{x}

+ x^{x}

+ x^{x - 1}

{\ln (x)}^{2} \cdot x^{x}

+ 2 \cdot \ln (x) \cdot x^{x}

+ x^{x}

+ x^{x - 1}

▶Derivative:

x^{x} \cdot {\ln (x)}^{2}

+ 2 x^{x} \cdot \ln (x)

+ x^{x}

+ x^{x - 1}