(\partial)/(\partial x)(5^{8^{x^2}})

Lösung

\frac{\partial}{\partial x} (5^{8^{x^{2}}})

3 ln (5) ln (2) \cdot 5^{8^{x^{2}}} \cdot 2^{3 x^{2} + 1} x

Schritte zur Lösung

\frac{\partial}{\partial x} (5^{8^{x^{2}}})

Wende Exponentenregel an:

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

5^{8^{x^{2}}} = e^{8^{x^{2}} l n (5)}

= \frac{\partial}{\partial x} (e^{8^{x^{2}} l n (5)})

Wende die Kettenregel an:

e^{8^{x^{2}} l n (5)} \frac{\partial}{\partial x} (8^{x^{2}} ln (5))

= e^{8^{x^{2}} l n (5)} \frac{\partial}{\partial x} (8^{x^{2}} ln (5))

\frac{\partial}{\partial x} (8^{x^{2}} ln (5)) = 3 ln (5) ln (2) \cdot 2^{3 x^{2} + 1} x

= e^{8^{x^{2}} l n (5)} \cdot 3 ln (5) ln (2) \cdot 2^{3 x^{2} + 1} x

e^{8^{x^{2}} l n (5)} = 5^{8^{x^{2}}}

= 3 ln (5) ln (2) \cdot 5^{8^{x^{2}}} \cdot 2^{3 x^{2} + 1} x

\int (e^{x} + 1) \cdot (x - 4) d x

d er i v a t i v e f (x) = ln (8 x)

t an g e n t f (x) = 25 - x^{2}, (5, 0)

d er i v a t i v e \frac{x + 7}{3 x ^{2} e ^{x}}

y^{^{'}} = (x + y + 7)^{2}