derivative of (1+3x^{1/(2x)})

Lösung

\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}}

Schritte zur Lösung

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

Wende Exponentenregel an:

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)})

Wende die Kettenregel an:

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

t an g e n t x, at x = 9

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

s l o p e (10, 5), (1, 12)

in v erse l a pl a ce \frac{1}{s ^{3} + 1}