derivative y=(x^2+1)^x

解答

导数

y = (x^{2} + 1)^{x}

(ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1}) (x^{2} + 1)^{x}

求解步骤

\frac{d}{d x} ((x^{2} + 1)^{x})

使用指数法则:

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

(x^{2} + 1)^{x} = e^{x l n (x^{2} + 1)}

= \frac{d}{d x} (e^{x l n (x^{2} + 1)})

使用连锁律:

e^{x l n (x^{2} + 1)} \frac{d}{d x} (x ln (x^{2} + 1))

= e^{x l n (x^{2} + 1)} \frac{d}{d x} (x ln (x^{2} + 1))

\frac{d}{d x} (x ln (x^{2} + 1)) = ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1}

= e^{x l n (x^{2} + 1)} (ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1})

e^{x l n (x^{2} + 1)} = (x^{2} + 1)^{x}

= (ln (x^{2} + 1) + \frac{2 x ^{2}}{x ^{2} + 1}) (x^{2} + 1)^{x}

\frac{d}{d x} (\frac{5}{x ^{7}})

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