In this post, we will learn how to find the derivative of the following function with respect to \( x \): \( \displaystyle f(x) = \ln\left(\frac{x^2 \sqrt{x^3 + 1}}{(2x + 1)^4}\right) \)
Solution:
\( \displaystyle f(x) = \ln\left(\frac{x^2 \sqrt{x^3 + 1}}{(2x + 1)^4}\right) \\ \displaystyle \Rightarrow f(x) = \ln(x^2) + \ln(\sqrt{x^3 + 1})\, – \ln((2x + 1)^4) \small \text{, using the properties of logarithms} \\ \displaystyle \Rightarrow f(x) = 2\ln(x) + \frac{1}{2}\ln(x^3 + 1)\, – 4\ln(2x + 1) \small \text{, using the properties of logarithms} \\ \displaystyle \Rightarrow \frac{d}{dx}f(x) = \frac{d}{dx}\left[2\ln(x)\right] + \frac{d}{dx}\left[\frac{1}{2}\ln(x^3 + 1)\right] – \frac{d}{dx}\left[4\ln(2x + 1)\right] \small \text{, differentiate each term w.r.t } x\\ \displaystyle \Rightarrow \frac{d}{dx}f(x) = \frac{2}{x} + \frac{1}{2} \cdot \frac{3x^2}{x^3 + 1}\, – 4 \cdot \frac{2}{2x + 1} \\ \displaystyle \Rightarrow \frac{d}{dx}f(x) = \frac{2}{x} + \frac{3x^2}{2(x^3 + 1)} – \frac{8}{2x + 1} \\ \)Therefore, the derivative of the function with respect to \( x \) is:
\[ \fbox{$ f'(x) = \frac{2}{x} + \frac{3x^2}{2(x^3 + 1)} – \frac{8}{2x + 1} $} \\ \]
Please let me know in the comments if you find any error in this solution.