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

Apr 30, 2016

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

#### Explanation:

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

• $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$
• $\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$
• $\frac{d}{\mathrm{dx}} x = 1$

we have

$\frac{d}{\mathrm{dx}} {x}^{7 x} = \frac{d}{\mathrm{dx}} {e}^{\ln \left({x}^{7 x}\right)}$

$= \frac{d}{\mathrm{dx}} {e}^{7 x \ln \left(x\right)}$

$= {e}^{7 x \ln \left(x\right)} \left(\frac{d}{\mathrm{dx}} 7 x \ln \left(x\right)\right)$

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

$= 7 {e}^{\ln \left({x}^{7 x}\right)} \left(x \frac{d}{\mathrm{dx}} \ln \left(x\right) + \ln \left(x\right) \frac{d}{\mathrm{dx}} x\right)$

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

$= 7 {x}^{7 x} \left(x \cdot \frac{1}{x} + \ln \left(x\right) \cdot 1\right)$

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