# Derivative of $x^{3 x}$

The calculator will find the derivative of $x^{3 x}$ using the logarithmic differentiation, with steps shown.

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Find $\frac{d}{dx} \left(x^{3 x}\right)$.

### Solution

Let $H{\left(x \right)} = x^{3 x}$.

Take the logarithm of both sides: $\ln\left(H{\left(x \right)}\right) = \ln\left(x^{3 x}\right)$.

Rewrite the RHS using the properties of logarithms: $\ln\left(H{\left(x \right)}\right) = 3 x \ln\left(x\right)$.

Differentiate separately both sides of the equation: $\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(3 x \ln\left(x\right)\right)$.

Differentiate the LHS of the equation.

The function $\ln\left(H{\left(x \right)}\right)$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = \ln\left(u\right)$ and $g{\left(x \right)} = H{\left(x \right)}$.

Apply the chain rule $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(H{\left(x \right)}\right)\right)}$$

The derivative of the natural logarithm is $\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(H{\left(x \right)}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(H{\left(x \right)}\right)$$

$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(H{\left(x \right)}\right)}}$$

Thus, $\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$.

Differentiate the RHS of the equation.

Apply the constant multiple rule $\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$ with $c = 3$ and $f{\left(x \right)} = x \ln\left(x\right)$:

$${\color{red}\left(\frac{d}{dx} \left(3 x \ln\left(x\right)\right)\right)} = {\color{red}\left(3 \frac{d}{dx} \left(x \ln\left(x\right)\right)\right)}$$

Apply the product rule $\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$ with $f{\left(x \right)} = x$ and $g{\left(x \right)} = \ln\left(x\right)$:

$$3 {\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = 3 {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:

$$3 x \frac{d}{dx} \left(\ln\left(x\right)\right) + 3 \ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 3 x \frac{d}{dx} \left(\ln\left(x\right)\right) + 3 \ln\left(x\right) {\color{red}\left(1\right)}$$

The derivative of the natural logarithm is $\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$:

$$3 x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + 3 \ln\left(x\right) = 3 x {\color{red}\left(\frac{1}{x}\right)} + 3 \ln\left(x\right)$$

Thus, $\frac{d}{dx} \left(3 x \ln\left(x\right)\right) = 3 \ln\left(x\right) + 3$.

Hence, $\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = 3 \ln\left(x\right) + 3$.

Therefore, $\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(3 \ln\left(x\right) + 3\right) H{\left(x \right)} = 3 x^{3 x} \left(\ln\left(x\right) + 1\right)$.

$\frac{d}{dx} \left(x^{3 x}\right) = 3 x^{3 x} \left(\ln\left(x\right) + 1\right)$A