derivative of 2^{x^3+3x}

解答

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

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

求解步骤

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

使用指数法则:

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

2^{x^{3} + 3 x} = e^{(x^{3} + 3 x) l n (2)}

= \frac{d}{d x} (e^{(x^{3} + 3 x) l n (2)})

使用连锁律:

e^{(x^{3} + 3 x) l n (2)} \frac{d}{d x} ((x^{3} + 3 x) ln (2))

= e^{(x^{3} + 3 x) l n (2)} \frac{d}{d x} ((x^{3} + 3 x) ln (2))

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

= e^{(x^{3} + 3 x) l n (2)} ln (2) (3 x^{2} + 3)

e^{(x^{3} + 3 x) l n (2)} = 2^{x^{3} + 3 x}

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

tan (t) y^{^{'}} - y = 1

\int x^{2} x^{2} - 9 d x

y^{^{''}} + 4 y = 10 x (4 - x), y (0) = - 5, y (4) = 10

\frac{d}{d t} (t^{3} + 3 t)

\frac{\partial}{\partial x} (y tan (x))