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The equation I'm trying to differentiate is, $ f(x) = \sqrt\frac{x^2-1}{x^2+1}$ and I know the answer is meant to be

$$=\frac{\frac{x\sqrt {x^2+1}}{\sqrt {x^2-1}}-\frac{x\sqrt {x^2-1}}{\sqrt {x^2+1}}}{x^2+1}$$

But when I do the working out I get this

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

$$=\frac{\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot\frac{1}{2}(x^2+1)^\frac{-1}{2}\cdot2x}{x^2+1}$$

simplify $$=\frac{x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot x(x^2+1)^\frac{-1}{2}}{x^2+1}$$

$$=\frac{\frac{\sqrt {x^2+1}}{x\sqrt {x^2-1}}-\frac{\sqrt {x^2-1}}{x\sqrt {x^2+1}}}{x^2+1}$$

As you can see two of my $x$'s are in the wrong location, and I just can't figure out what I'm doing wrong. Any help as to what steps I'm doing wrong or missing would be much appreciated.

Asaf Karagila
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DrMolo
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    Your last step looks as if it misplaces two $x$s – Henry Aug 13 '19 at 01:40
  • my reasoning for putting the two $x$'s on the bottom is because they're apart of $x(x^2-1)^\frac{-1}{2}$ and $x(x^2+1)^\frac{-1}{2}$, and both have negative exponents. – DrMolo Aug 13 '19 at 01:44
  • @DrMolo As I explained in my updated answer, the $\frac{-1}{2}$ powers only apply to the expressions in the brackets, i.e., the $x^2 - 1$ and $x^2 + 1$. As the $x$ factors are outside of the brackets, they are not affected by that. For example, $x(x^2 - 1)^{\frac{-1}{2}} = x\left((x^2 - 1)^{\frac{-1}{2}}\right)$. – John Omielan Aug 13 '19 at 01:49

3 Answers3

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Going from your second last line to the last line, you put the $x$ factors into the denominator. Instead, you need to leave them in the numerator. This is because the power of $\frac{-1}{2}$ only apply to the expressions in the brackets, but the $x$ factors are outside of the brackets, so they are not affected. Thus, for example, with the first term in the numerator, you have

$$\begin{equation}\begin{aligned} x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2} & = x\left((x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}\right) \\ & = x\left(\frac{\sqrt {x^2+1}}{\sqrt {x^2-1}}\right) \\ & = \frac{x\sqrt {x^2+1}}{\sqrt {x^2-1}} \end{aligned}\end{equation}\tag{1}\label{eq1}$$

John Omielan
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1

You have erroneously transposed the two lone $x$'s in the second-last line into $1/x$'s in the last. (That is, they incongrously turn from being mulitplied with the radicals to dividing them.)

$$=\frac{\color{blue}x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot \color{blue}x(x^2+1)^\frac{-1}{2}}{x^2+1}$$

$$=\frac{\frac{\sqrt {x^2+1}}{\color{red}x\sqrt {x^2-1}}-\frac{\sqrt {x^2-1}}{\color{red}x\sqrt {x^2+1}}}{x^2+1}$$

Parcly Taxel
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0

Just a small trick.

When you face expressions which contains products, quotients, powers, logarithmic differentiation makes life easier.

$$y= \sqrt\frac{x^2-1}{x^2+1}\implies \log(y)=\frac 12 \log(x^2-1)-\frac 12 \log(x^2+1)$$ Differentiate both sides $$\frac {y'} y=\frac 12 \frac {2x}{x^2-1}-\frac 12 \frac {2x}{x^2+1}=\frac{2x}{(x^2-1)(x^2+1)}$$

$$y'=y\times\frac {y'} y=\sqrt\frac{x^2-1}{x^2+1}\times\frac{2x}{(x^2-1)(x^2+1)}$$ and simplify.

Claude Leibovici
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