# Derivative of $\left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}$

The calculator will find the derivative of $\left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}$ using the logarithmic differentiation, with steps shown.

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### Your Input

Find $\frac{d}{dx} \left(\left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}\right)$.

### Solution

Let $H{\left(x \right)} = \left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}$.

Take the logarithm of both sides: $\ln\left(H{\left(x \right)}\right) = \ln\left(\left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}\right)$.

Rewrite the RHS using the properties of logarithms: $\ln\left(H{\left(x \right)}\right) = 4 \ln\left(x^{3} + 4\right) + 2 \ln\left(x^{5} + 2\right)$.

Differentiate separately both sides of the equation: $\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right) + 2 \ln\left(x^{5} + 2\right)\right)$.

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 $\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$:

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

Return to the old variable:

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

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.

The derivative of a sum/difference is the sum/difference of derivatives:

$${\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right) + 2 \ln\left(x^{5} + 2\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right)\right) + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right)\right)}$$

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 $c = 4$ and $f{\left(x \right)} = \ln\left(x^{3} + 4\right)$:

$${\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right)\right)\right)} + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) = {\color{red}\left(4 \frac{d}{dx} \left(\ln\left(x^{3} + 4\right)\right)\right)} + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right)$$

The function $\ln\left(x^{3} + 4\right)$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = \ln\left(u\right)$ and $g{\left(x \right)} = x^{3} + 4$.

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

$$4 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{3} + 4\right)\right)\right)} + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) = 4 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{3} + 4\right)\right)} + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right)$$

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

$$4 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{3} + 4\right) + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) = 4 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{3} + 4\right) + \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right)$$

Return to the old variable:

$$\frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) + \frac{4 \frac{d}{dx} \left(x^{3} + 4\right)}{{\color{red}\left(u\right)}} = \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) + \frac{4 \frac{d}{dx} \left(x^{3} + 4\right)}{{\color{red}\left(x^{3} + 4\right)}}$$

The derivative of a sum/difference is the sum/difference of derivatives:

$$\frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) + \frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{3} + 4\right)\right)}}{x^{3} + 4} = \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) + \frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) + \frac{d}{dx} \left(4\right)\right)}}{x^{3} + 4}$$

The derivative of a constant is $0$:

$$\frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) + \frac{4 \left({\color{red}\left(\frac{d}{dx} \left(4\right)\right)} + \frac{d}{dx} \left(x^{3}\right)\right)}{x^{3} + 4} = \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) + \frac{4 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{3}\right)\right)}{x^{3} + 4}$$

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 3$:

$$\frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) + \frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)}}{x^{3} + 4} = \frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right) + \frac{4 {\color{red}\left(3 x^{2}\right)}}{x^{3} + 4}$$

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 $c = 2$ and $f{\left(x \right)} = \ln\left(x^{5} + 2\right)$:

$$\frac{12 x^{2}}{x^{3} + 4} + {\color{red}\left(\frac{d}{dx} \left(2 \ln\left(x^{5} + 2\right)\right)\right)} = \frac{12 x^{2}}{x^{3} + 4} + {\color{red}\left(2 \frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right)\right)}$$

The function $\ln\left(x^{5} + 2\right)$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = \ln\left(u\right)$ and $g{\left(x \right)} = x^{5} + 2$.

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

$$\frac{12 x^{2}}{x^{3} + 4} + 2 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{5} + 2\right)\right)\right)} = \frac{12 x^{2}}{x^{3} + 4} + 2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{5} + 2\right)\right)}$$

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

$$\frac{12 x^{2}}{x^{3} + 4} + 2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{5} + 2\right) = \frac{12 x^{2}}{x^{3} + 4} + 2 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{5} + 2\right)$$

Return to the old variable:

$$\frac{12 x^{2}}{x^{3} + 4} + \frac{2 \frac{d}{dx} \left(x^{5} + 2\right)}{{\color{red}\left(u\right)}} = \frac{12 x^{2}}{x^{3} + 4} + \frac{2 \frac{d}{dx} \left(x^{5} + 2\right)}{{\color{red}\left(x^{5} + 2\right)}}$$

The derivative of a sum/difference is the sum/difference of derivatives:

$$\frac{12 x^{2}}{x^{3} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{5} + 2\right)\right)}}{x^{5} + 2} = \frac{12 x^{2}}{x^{3} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{5}\right) + \frac{d}{dx} \left(2\right)\right)}}{x^{5} + 2}$$

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 5$:

$$\frac{12 x^{2}}{x^{3} + 4} + \frac{2 \left({\color{red}\left(\frac{d}{dx} \left(x^{5}\right)\right)} + \frac{d}{dx} \left(2\right)\right)}{x^{5} + 2} = \frac{12 x^{2}}{x^{3} + 4} + \frac{2 \left({\color{red}\left(5 x^{4}\right)} + \frac{d}{dx} \left(2\right)\right)}{x^{5} + 2}$$

The derivative of a constant is $0$:

$$\frac{12 x^{2}}{x^{3} + 4} + \frac{2 \left(5 x^{4} + {\color{red}\left(\frac{d}{dx} \left(2\right)\right)}\right)}{x^{5} + 2} = \frac{12 x^{2}}{x^{3} + 4} + \frac{2 \left(5 x^{4} + {\color{red}\left(0\right)}\right)}{x^{5} + 2}$$

Thus, $\frac{d}{dx} \left(4 \ln\left(x^{3} + 4\right) + 2 \ln\left(x^{5} + 2\right)\right) = \frac{10 x^{4}}{x^{5} + 2} + \frac{12 x^{2}}{x^{3} + 4}$.

Hence, $\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = \frac{10 x^{4}}{x^{5} + 2} + \frac{12 x^{2}}{x^{3} + 4}$.

Therefore, $\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(\frac{10 x^{4}}{x^{5} + 2} + \frac{12 x^{2}}{x^{3} + 4}\right) H{\left(x \right)} = 2 x^{2} \left(x^{3} + 4\right)^{3} \left(x^{5} + 2\right) \left(11 x^{5} + 20 x^{2} + 12\right).$

### Answer

$\frac{d}{dx} \left(\left(x^{3} + 4\right)^{4} \left(x^{5} + 2\right)^{2}\right) = 2 x^{2} \left(x^{3} + 4\right)^{3} \left(x^{5} + 2\right) \left(11 x^{5} + 20 x^{2} + 12\right)$A