We are given the following function to differentiate:

$y = \log_3\left(\left(\frac{3x^4}{5x^3 + 3}\right)^3\right)$

To find the derivative of this function, we can follow these steps:

Step 1: Apply the change of base formula for logarithms

We can simplify the expression by changing the base of the logarithm from base 3 to base $e$. Using the change of base formula: $\log_b(x) = \frac{\ln(x)}{\ln(b)}$ we rewrite the given function as: $y = \frac{\ln\left(\left(\frac{3x^4}{5x^3 + 3}\right)^3\right)}{\ln(3)}$

Step 2: Simplify the logarithmic expression

Now, apply the logarithm power rule, $\ln(a^b) = b \ln(a)$, to bring the exponent 3 in front: $y = \frac{3 \ln\left(\frac{3x^4}{5x^3 + 3}\right)}{\ln(3)}$

Step 3: Apply the logarithm quotient rule

Next, apply the logarithmic rule for division, $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$: $y = \frac{3}{\ln(3)} \left[\ln(3x^4) - \ln(5x^3 + 3)\right]$

Step 4: Differentiate each term

Now, differentiate the expression term by term. Let’s differentiate each logarithmic term using the chain rule.

• For $\ln(3x^4)$: $\frac{d}{dx} \ln(3x^4) = \frac{1}{3x^4} \cdot \frac{d}{dx}(3x^4) = \frac{1}{3x^4} \cdot 12x^3 = \frac{4}{x}$

• For $\ln(5x^3 + 3)$: $\frac{d}{dx} \ln(5x^3 + 3) = \frac{1}{5x^3 + 3} \cdot \frac{d}{dx}(5x^3 + 3) = \frac{1}{5x^3 + 3} \cdot 15x^2 = \frac{15x^2}{5x^3 + 3}$

Step 5: Put everything together

The derivative of the entire function is: $\frac{dy}{dx} = \frac{3}{\ln(3)} \left(\frac{4}{x} - \frac{15x^2}{5x^3 + 3}\right)$

This is the derivative of the given function.

Would you like further details or have any questions on these steps?

Related Questions:

1. How do you apply the chain rule to logarithmic functions?
2. What is the change of base formula and how is it used in differentiation?
3. How can you differentiate a function involving a quotient of polynomials?
4. What is the logarithmic power rule and when should it be used?
5. How do you simplify complex logarithmic expressions before differentiating?

Tip:

Always simplify the logarithmic expression as much as possible before differentiating, as this can make the derivative calculation easier.