derivative of 5^{x+1}

解答

\frac{d}{d x} (5^{x + 1})

ln (5) \cdot 5^{x + 1}

求解步骤

\frac{d}{d x} (5^{x + 1})

使用指数法则:

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

5^{x + 1} = e^{(x + 1) l n (5)}

= \frac{d}{d x} (e^{(x + 1) l n (5)})

使用连锁律:

e^{(x + 1) l n (5)} \frac{d}{d x} ((x + 1) ln (5))

= e^{(x + 1) l n (5)} \frac{d}{d x} ((x + 1) ln (5))

\frac{d}{d x} ((x + 1) ln (5)) = ln (5)

= e^{(x + 1) l n (5)} ln (5)

e^{(x + 1) l n (5)} = 5^{x + 1}

= ln (5) \cdot 5^{x + 1}

t an g e n t - 3 x^{2}

\frac{d}{d x} (\frac{4}{4 x + 1})

\int \frac{4 + 8 x}{1 + x ^{2}} d x

\int 9 x^{2} + 4 x - 3 d x

\frac{d}{d x} (\frac{2 x - x ^{2}}{x} - 3 cot (x))