derivative of (θ(2xy(θ,x,d)^3)dx)

解答

\frac{d}{d x} ((θ) (2 x y (θ, x, d) (θ, x, d)^{3}) d x)

θ \cdot 2 d (1 \cdot y (θ, x, d) (θ, x, d)^{3} x + (\frac{d}{d x} (y (θ, x, d)) (θ, x, d)^{3} x + (3 (θ, x, d)^{2} (θ,, d \cdot 1) x + 1 \cdot (θ, x, d)^{3}) y (θ, x, d)) x)

求解步骤

\frac{d}{d x} ((θ) (2 x y (θ, x, d) (θ, x, d)^{3}) d x)

将

θ, d

视为常数

将常数提出:

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

= θ \cdot 2 d \frac{d}{d x} (x y (θ, x, d) (θ, x, d)^{3} x)

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = y (θ, x, d) (θ, x, d)^{3} x

= θ \cdot 2 d (\frac{d x}{d x} y (θ, x, d) (θ, x, d)^{3} x + \frac{d}{d x} (y (θ, x, d) (θ, x, d)^{3} x) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (y (θ, x, d) (θ, x, d)^{3} x) = \frac{d}{d x} (y (θ, x, d)) (θ, x, d)^{3} x + (3 (θ, x, d)^{2} (θ,, d \cdot 1) x + 1 \cdot (θ, x, d)^{3}) y (θ, x, d)

= θ \cdot 2 d (1 \cdot y (θ, x, d) (θ, x, d)^{3} x + (\frac{d}{d x} (y (θ, x, d)) (θ, x, d)^{3} x + (3 (θ, x, d)^{2} (θ,, d \cdot 1) x + 1 \cdot (θ, x, d)^{3}) y (θ, x, d)) x)

f (x) = - 2 + 3 e^{(- c o s (x))}

s im pl i f y a \cdot \frac{d}{d x} (y (x)^{2})

f a c t or - 3 x^{2} + \cdot (- 4)

s im pl i f y [- 2.6]

f a c t or W (e^{2})