(\partial)/(\partial x)(2^{x^2-y^5})

解答

\frac{\partial}{\partial x} (2^{x^{2} - y^{5}})

ln (2) \cdot 2^{x^{2} - y^{5} + 1} x

求解步骤

\frac{\partial}{\partial x} (2^{x^{2} - y^{5}})

将

y

视为常数

使用指数法则:

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

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

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

使用连锁律:

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

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

\frac{\partial}{\partial x} ((x^{2} - y^{5}) ln (2)) = 2 ln (2) x

= e^{(x^{2} - y^{5}) l n (2)} \cdot 2 ln (2) x

化简

e^{(x^{2} - y^{5}) l n (2)} \cdot 2 ln (2) x : ln (2) \cdot 2^{x^{2} - y^{5} + 1} x

= ln (2) \cdot 2^{x^{2} - y^{5} + 1} x

t an g e n t x^{3} - 8 x^{2} + 5 x - 2

\frac{\partial}{\partial x} (4 e^{x y + 3})

d er i v a t i v e (7 x^{6} + 2 x^{3})^{4}

t an g e n t x + 8

n = 0 \sum \infty 3^{n} 7^{- n}