d/(d{x)}(({x}^2+{y}^2+{z}^2)^{1/2})

Lösung

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

\frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{1}{2}}}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

= \frac{1}{2 ( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{1}{2}}} \frac{d}{d x} (x^{2} + y^{2} + z^{2})

\frac{d}{d x} (x^{2} + y^{2} + z^{2}) = 2 x

= \frac{1}{2 ( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{1}{2}}} \cdot 2 x

Vereinfache

\frac{1}{2 ( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{1}{2}}} \cdot 2 x : \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{1}{2}}}

\frac{d}{d x} (6 ln (sec (x) + tan (x)))

d er i v a t i v e cos^{3} (x)

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{15}}

d er i v a t i v e - 6 sin (t)

d er i v a t i v e ln (e^{x} + x)