How do you find the derivative of $x^{7 x}$ $x^(7x)$ ?

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Answer 1 · 2016-04-30T22:33:25

$\frac{d}{d x} x^{7 x} = 7 x^{7 x} (ln (x) + 1)$ $d/dx x^(7x)=7x^(7x)(ln(x)+1)$

Using the chain rule and the product rule, together with the following derivatives:

we have

$\frac{d}{d x} x^{7 x} = \frac{d}{d x} e^{ln (x^{7 x})}$ $d/dx x^(7x) = d/dx e^(ln(x^(7x)))$

$= \frac{d}{d x} e^{7 x ln (x)}$ $=d/dx e^(7xln(x))$

$= e^{7 x ln (x)} (\frac{d}{d x} 7 x ln (x))$ $=e^(7xln(x))(d/dx7xln(x))$

(by the chain rule with the functions $e^{x}$ $e^x$ and $7 x ln (x)$ $7xln(x)$ )

$= 7 e^{ln (x^{7 x})} (x \frac{d}{d x} ln (x) + ln (x) \frac{d}{d x} x)$ $=7e^(ln(x^(7x)))(xd/dxln(x) + ln(x)d/dxx)$

(by the product rule, and factoring out the $7$ $7$ )

$= 7 x^{7 x} (x \cdot \frac{1}{x} + ln (x) \cdot 1)$ $=7x^(7x)(x*1/x + ln(x)*1)$

$= 7 x^{7 x} (ln (x) + 1)$ $=7x^(7x)(ln(x)+1)$

How do you find the derivative of x7xx^(7x)?