What is the derivative of (1+3x^{1/(2x)}) ?

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

What is the first derivative of (1+3x^{1/(2x)}) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

(1 + 3 x)^{\frac{1}{2 x}} = e^{\frac{1}{2 x} l n (1 + 3 x)}

= \frac{d}{d x} (e^{\frac{1}{2 x} l n (1 + 3 x)})

Apply the chain rule:

e^{\frac{1}{2 x} l n (1 + 3 x)} \frac{d}{d x} (\frac{1}{2 x} ln (1 + 3 x))

= e^{\frac{1}{2 x} l n (1 + 3 x)} \frac{d}{d x} (\frac{1}{2 x} ln (1 + 3 x))

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

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

e^{\frac{1}{2 x} l n (1 + 3 x)} = (1 + 3 x)^{\frac{1}{2 x}}

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

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} (\frac{d y}{d t}) + \frac{t}{2 y ^{4}} = 0

x, at x = 9

\frac{\partial}{\partial y} (4 y)

(10, 5), (1, 12)

\frac{1}{s ^{3} + 1}

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