Respuesta :
Solution
Write the function
[tex]f(x)\text{ = }\ln\left(\frac{x^2\sin\left(x\right)}{\arctan\left(x\right)}\right)[/tex]Next
Apply the chain rule
[tex]\begin{gathered} Let\text{ u =}\left(\frac{x^2\sin\left(x\right)}{\arctan\left(x\right)}\right) \\ \mathrm{Apply\:the\:Quotient\:Rule}:\quad \left(\frac{f}{g}\right)^'=\frac{f\:'\cdot g-g'\cdot f}{g^2} \\ =\frac{\frac{d}{dx}\left(x^2\sin \left(x\right)\right)\arctan \left(x\right)-\frac{d}{dx}\left(\arctan \left(x\right)\right)x^2\sin \left(x\right)}{\left(\arctan \left(x\right)\right)^2} \\ \frac{d}{dx}\left(x^2\sin\left(x\right)\right)=\frac{d}{dx}\left(x^2\sin\left(x\right)\right) \\ \\ \frac{d}{dx}\left(\arctan\left(x\right)\right)\text{ = }\frac{1}{x^2+1} \\ \\ =\frac{\left(2x\sin \left(x\right)+\cos \left(x\right)x^2\right)\arctan \left(x\right)-\frac{1}{x^2+1}x^2\sin \left(x\right)}{\left(\arctan \left(x\right)\right)^2} \\ \\ Simplify \\ \\ \frac{du}{df}=\frac{\arctan\left(x\right)\left(2x\sin\left(x\right)+x^2\cos\left(x\right)\right)\left(x^2+1\right)-x^2\sin\left(x\right)}{\arctan^2\left(x\right)\left(x^2+1\right)} \end{gathered}[/tex]Next
[tex]\begin{gathered} f(x)\text{ = In\lparen u\rparen} \\ \frac{df}{du}\text{ = }\frac{1}{u} \\ \\ From\text{ chain rule} \\ \\ f^{\prime}(x)=\text{ }\frac{du}{dx}\text{ }\times\text{ }\frac{df}{du} \\ \\ =\frac{1}{\frac{x^2\sin\left(x\right)}{\arctan\left(x\right)}}\frac{d}{dx}\left(\frac{x^2\sin\left(x\right)}{\arctan\left(x\right)}\right) \\ \\ =\frac{1}{\frac{x^2\sin \left(x\right)}{\arctan \left(x\right)}}\cdot \frac{\arctan \left(x\right)\left(2x\sin \left(x\right)+x^2\cos \left(x\right)\right)\left(x^2+1\right)-x^2\sin \left(x\right)}{\arctan ^2\left(x\right)\left(x^2+1\right)} \\ \\ =\frac{\arctan \left(x\right)\left(x^2\cos \left(x\right)+2x\sin \left(x\right)\right)\left(x^2+1\right)-x^2\sin \left(x\right)}{x^2\sin \left(x\right)\arctan \left(x\right)\left(x^2+1\right)} \\ \\ \end{gathered}[/tex]Final answer
The derivative of the function is given below:
[tex]f^{\prime}(x)\text{ }=\frac{\arctan\left(x\right)\left(x^2\cos\left(x\right)+2x\sin\left(x\right)\right)\left(x^2+1\right)-x^2\sin\left(x\right)}{x^2\sin\left(x\right)\arctan\left(x\right)\left(x^2+1\right)}[/tex]