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Finding the derivative of $$y=\frac{\sqrt{2x^2}}{\cos x}$$ I am going through the steps and having trouble using the quotient rule. I have seen the final answer, and I've had no trouble using the quotient rule in the past, but this one is giving me trouble in terms of figuring out where all the fractions end up going. Below is what I have so far:

$$\frac{dy}{dx}=\frac{\dfrac{4x\cos x}{2\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}$$

I assume I can get rid of the 2 in the denominator in the first term in the numerator, leaving $\dfrac{2x\cos x}{\sqrt{2x^2}}$ and after seeing the final answer, I believe the denominator can be moved down so that the final answer's denominator is $\sqrt{2x^2}(\cos x)^2$. However, in the final answer the numerator is $2x\cos x +2x^2\sin x$. Where does the $2x^2$ come from? How could the square root have gone away? For reference, here is the final answer I'm supposed to get:

$$\frac{2x\cos x+2x^2\sin x}{\sqrt{2x^2}(\cos x)^2}$$

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    $\begingroup$ Multiply numerator and denominator by $\sqrt {2x^{2}}$. $\endgroup$ Jun 26, 2020 at 7:34
  • $\begingroup$ Yes, that is suddenly obvious to me. Thank you for mentioning it, for some reason I just didn't realize it. Very simple! $\endgroup$
    – walyba
    Jun 26, 2020 at 7:39
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    $\begingroup$ Multiply top and bottom by $\sqrt{2x^2}$. Bigger question is the first power of cosine they have in the denominator. $\endgroup$
    – Mike
    Jun 26, 2020 at 7:40

1 Answer 1

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You have $$\frac{\frac{4x\cos x}{2\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}$$ You correctly mentioned that this can be simplified to $$\frac{\frac{2x\cos x}{\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}$$ To simplify this you can remove the compound fraction. To get rid of the fraction in the numerator, multiply numerator and denominator of the big fraction by the denominator of the small fraction. That is, multiply the big fraction by $\frac{\sqrt{2x^2}}{\sqrt{2x^2}}$, giving: $$\frac{\frac{2x\cos x}{\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}\times\frac{\sqrt{2x^2}}{\sqrt{2x^2}}$$ The $\sqrt{2x^2}$ term in the denominator of the fraction in the numerator is cancelled, and $\sqrt{2x^2}\times\sqrt{2x^2}=2x^2$, this is how the square root disappears. So it simplifies to $$\frac{2x\cos x+2x^2\sin x}{\sqrt{2x^2}(\cos x)^2}$$ as you mentioned.

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