find the derivative of the function y = (cos³(x)sin³(x))/((2x³ + 1)²) using logarithmic differentiation. be…

Answer:

$y'\left(x\right)=\frac{\cos^{3}(x)\sin^{3}(x)}{(2x^{3}+1)^{2}}\left(3\tan(x)-3\cot(x)-\frac{12x^{2}}{2x^{3}+1}\right)$

Explanation:

Step1: Take natural - log of both sides

$\ln(y)=\ln\left(\frac{\cos^{3}(x)\sin^{3}(x)}{(2x^{3}+1)^{2}}\right)$ Using the properties of logarithms $\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$ and $\ln(ab)=\ln(a)+\ln(b)$, we get $\ln(y)=3\ln(\cos(x)) + 3\ln(\sin(x))-2\ln(2x^{3}+1)$.

Step2: Differentiate both sides with respect to x

$\frac{y'}{y}=3\frac{-\sin(x)}{\cos(x)}+3\frac{\cos(x)}{\sin(x)}-2\frac{6x^{2}}{2x^{3}+1}$

Step3: Simplify the right - hand side

$\frac{y'}{y}=- 3\tan(x)+3\cot(x)-\frac{12x^{2}}{2x^{3}+1}$

Step4: Solve for y'

$y' = y\left(-3\tan(x)+3\cot(x)-\frac{12x^{2}}{2x^{3}+1}\right)$ Substitute $y = \frac{\cos^{3}(x)\sin^{3}(x)}{(2x^{3}+1)^{2}}$ into the above equation: $y'=\frac{\cos^{3}(x)\sin^{3}(x)}{(2x^{3}+1)^{2}}\left(3\tan(x)-3\cot(x)-\frac{12x^{2}}{2x^{3}+1}\right)$