Differentiating product and squares of logarithms

Question

I need help differentiating. I am really confused how to solve with the $\ln x$ in the equation. Which of the logarithm rules do I need to use for this equation?

$$y= 12x \ln x + 12x - 6x (\ln x)^2 + 8$$

Answer 1

To find the derivative of $12x\ln x$, use the product rule to get $$12\ln x + 12x(1/x) = 12(\ln x +1)$$

To find the derivative of $(6x)(\ln x)^2$, use the product rule and the chain rule to get $$6(\ln x)^2 + 6x(2)(\ln x)(1/x) = 6\ln x((\ln x) + 2)$$

Answer 2

You wouldn't use any of the normal logarithmic rules, you'd instead use the product and chain rule. Namely, you get $$\frac{\mathrm{d}y}{\mathrm{d}x} = 12\ln x + 12x \cdot \frac{1}{x} + 12 - 6 (\ln x)^2 - 6x \cdot 2 \ln x \cdot \frac{1}{x}$$

This simplifies to $$\frac{\mathrm{d}y}{\mathrm{d}x} = 24 - 6(\ln x)^2$$

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The product rule states that for a product of function $uv$, the derivative is given by $$\frac{\mathrm{d}}{\mathrm{d}x} uv = u \, \mathrm{d}v + v\, \mathrm{d}u$$

The chain rule states that for a composition of function $f(g(x))$, the derivative is given by $$\frac{\mathrm{d}}{\mathrm{d}x} f(g(x)) = g'(x)f'(g(x))$$