derivative of (x+1^{sin(x)})

解答

\frac{d}{d x} ((x + 1)^{s i n (x)})

(cos (x) ln (x + 1) + \frac{sin ( x )}{x + 1}) (x + 1)^{s i n (x)}

求解步骤

\frac{d}{d x} ((x + 1)^{s i n (x)})

使用指数法则:

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

(x + 1)^{s i n (x)} = e^{s i n (x) l n (x + 1)}

= \frac{d}{d x} (e^{s i n (x) l n (x + 1)})

使用连锁律:

e^{s i n (x) l n (x + 1)} \frac{d}{d x} (sin (x) ln (x + 1))

= e^{s i n (x) l n (x + 1)} \frac{d}{d x} (sin (x) ln (x + 1))

\frac{d}{d x} (sin (x) ln (x + 1)) = cos (x) ln (x + 1) + \frac{sin ( x )}{x + 1}

= e^{s i n (x) l n (x + 1)} (cos (x) ln (x + 1) + \frac{sin ( x )}{x + 1})

e^{s i n (x) l n (x + 1)} = (x + 1)^{s i n (x)}

= (cos (x) ln (x + 1) + \frac{sin ( x )}{x + 1}) (x + 1)^{s i n (x)}

2 \frac{d y}{d x} = y - e^{4 x}

s l o p e f (x) = 6 x (6.6)

\frac{\partial}{\partial y} (x sin (x + 2 y))

y^{^{''}} - 2 y^{^{'}} - 8 y = 7 - e^{4 x}

t an g e n t f (x) = x^{3} + 9 x^{2} + 34 x + 15, at x = 7