derivative of xey(x^2+yln(x))

解答

\frac{d}{d x} (x e y (x) (x)^{2} + y (x) ln (x))

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

求解步骤

\frac{d}{d x} (x e y (x) (x)^{2} + y (x) ln (x))

化简

x e y (x) (x)^{2} + y (x) ln (x) : e x^{3} y (x) + ln (x) y (x)

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

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} (ln (x) y (x)) = \frac{y ( x )}{x} + ln (x) \frac{d}{d x} (y (x))

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

\int \frac{1}{x ^{4} 1 + x ^{2}} d x

t a y l or ln (1 - 3 y)

\int \frac{( x ^{2} )}{x ^{2} + 9} d x

\int \frac{tan ^{2} ( θ )}{sec ( θ )} d θ

- y + \frac{d y}{d x} ln (x) = 0