derivative of-10^{3x^2-4}

解答

\frac{d}{d x} (- 1 0^{3 x^{2} - 4})

- 3 ln (10) \cdot 5^{3 x^{2} - 4} \cdot 2^{3 x^{2} - 3} x

求解步骤

\frac{d}{d x} (- 1 0^{3 x^{2} - 4})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{d}{d x} (1 0^{3 x^{2} - 4})

使用指数法则:

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

1 0^{3 x^{2} - 4} = e^{(3 x^{2} - 4) l n (10)}

= - \frac{d}{d x} (e^{(3 x^{2} - 4) l n (10)})

使用连锁律:

e^{(3 x^{2} - 4) l n (10)} \frac{d}{d x} ((3 x^{2} - 4) ln (10))

= e^{(3 x^{2} - 4) l n (10)} \frac{d}{d x} ((3 x^{2} - 4) ln (10))

\frac{d}{d x} ((3 x^{2} - 4) ln (10)) = 6 ln (10) x

= - e^{(3 x^{2} - 4) l n (10)} \cdot 6 ln (10) x

化简

- e^{(3 x^{2} - 4) l n (10)} \cdot 6 ln (10) x : - 3 ln (10) \cdot 5^{3 x^{2} - 4} \cdot 2^{3 x^{2} - 3} x

= - 3 ln (10) \cdot 5^{3 x^{2} - 4} \cdot 2^{3 x^{2} - 3} x

\frac{d}{d x} (\frac{2}{4 x ^{3}})

x \frac{d y}{d x} = e^{y + x}

t an g e n t x^{2} + 2 x y - y^{2} + x = 27, (4, 7)

\frac{d y}{d x} = \frac{y ^{3}}{1 - 2 x y ^{2}}, y (0) = 1

\frac{\partial}{\partial x} (sin (z^{2}) cos (x y))