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Differentiate:

$$\ln\left(\dfrac{x^2\sqrt{2x^2+3}}{\left(x^4+x^2\right)^3}\right)$$

I have tried to figure it out here:

The steps are too long so I tidy up as an image

After the steps from the image, these are the final steps of simplifying: $$=\dfrac{\frac{2x^5\left(x^2+1\right)^3}{\sqrt{2x^2+3}}-\sqrt{2x^2+3}\left(4x^3\left(x^2+1\right)^3+6x^5\left(x^2+1\right)^2\right)}{x^4\left(x^2+1\right)^3\sqrt{2x^2+3}}$$

$$=\dfrac{x^4\left(x^2+1\right)^3\left(-\frac{4\sqrt{2x^2+3}}{x^5\left(x^2+1\right)^3}-\frac{6\sqrt{2x^2+3}}{x^3\left(x^2+1\right)^4}+\frac{2}{x^3\left(x^2+1\right)^3\sqrt{2x^2+3}}\right)}{\sqrt{2x^2+3}}$$

$$=\dfrac{2x}{2x^2+3}-\dfrac{6x}{x^2+1}-\dfrac{4}{x}$$

Please tell me whether this is correct or not, I would like to simplify my steps further if I could. Thanks!

nerdy
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1 Answers1

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To simplify we can use that

$$f(x)=\ln\left(\dfrac{x^2\sqrt{2x^2+3}}{\left(x^4+x^2\right)^3}\right)=\ln x^2+\ln\sqrt{2x^2+3}-\ln \left(x^4+x^2\right)^3=$$

$$=2\ln x+\frac12\ln(2x^2+3)-3\ln(x^4+x^2) $$

and therefore

$$f'(x)=2\cdot \frac1x+\frac12\cdot\frac{4x}{2x^2+3}-3\cdot\frac{4x^3+2x}{x^4+x^2}=\frac 2x+\frac{2x}{2x^2+3}-\frac{12x^2+6}{x(x^2+1)}$$

which is equivalent to the expression you have found indeed

$$\frac 2x+\frac{2x}{2x^2+3}-\frac{12x^2+6}{x(x^2+1)}=\frac{2x}{2x^2+3}-\frac{12x^2+6-2(x^2+1)}{x(x^2+1)}=$$

$$=\frac{2x}{2x^2+3}-\frac{10x^2+4}{x(x^2+1)}=\frac{2x}{2x^2+3}-\frac{4(x^2+1)}{x(x^2+1)}-\frac{6x^2}{x(x^2+1)}=$$

$$=\frac{2x}{2x^2+3}-\frac{4}{x}-\frac{6x}{x^2+1}$$

user
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