(\partial)/(\partial x)(({f}(x))^')

Lösung

\frac{\partial}{\partial x} ((f (x))^{^{'}})

' \frac{\partial}{\partial x} (f (x)) f (x)^{' - 1}

Schritte zur Lösung

\frac{\partial}{\partial x} ((f (x))^{^{'}})

Wende Exponentenregel an:

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

f (x)^{^{'}} = e^{^{'} l n (f (x))}

= \frac{\partial}{\partial x} (e^{^{^{'}} l n (f (x))})

Wende die Kettenregel an:

e^{^{'} l n (f (x))} \frac{\partial}{\partial x} (^{'} ln (f (x)))

= e^{^{'} l n (f (x))} \frac{\partial}{\partial x} (^{'} ln (f (x)))

\frac{\partial}{\partial x} (^{'} ln (f (x))) = \frac{^{'} \frac{\partial}{\partial x} ( f ( x ) )}{f ( x )}

= e^{^{^{'}} l n (f (x))} \frac{^{^{'}} \frac{\partial}{\partial x} ( f ( x ) )}{f ( x )}

Vereinfache

e^{^{'} l n (f (x))} \frac{^{'} \frac{\partial}{\partial x} ( f ( x ) )}{f ( x )} :^{'} \frac{\partial}{\partial x} (f (x)) f (x)^{^{'} - 1}

=^{'} \frac{\partial}{\partial x} (f (x)) f (x)^{^{'} - 1}

x y^{^{''}} + y^{^{'}} = x

a y^{^{'}} + b y = 0

4 t x \frac{d x}{d t} = x^{2} + 1

x \to 1 lim ((3 x - 4)^{2})

d er i v a t i v e f (x) = 5 - x^{2}