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Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$

My attempt at the question:

Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$ Differentiating with respect to x: $$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $$ Therefore

$$f'(g(x)) = \frac{1}{2(g'(x))\sqrt{x-1}} $$

Finding the value of $x$ for which $g(x) = 1$ : $ x=( -1) , x=2$ But as $x\neq (-1)$, as $\sqrt{x-1}$ becomes indeterminant, substitute x = 2. we get: $$f'(1) = \frac13 $$ Which is not the correct answer. The correct answer is supposedly $1$. Need some help as to why my method is wrong.

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    $\begingroup$ Sorry, I find no error. $\endgroup$
    – Shine
    Aug 3, 2014 at 13:57
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    $\begingroup$ maybe you should give the correct answer what the text says is? $\endgroup$
    – RE60K
    Aug 3, 2014 at 13:59
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    $\begingroup$ I also think the answer is 1/3. $\endgroup$
    – Mick A
    Aug 3, 2014 at 14:43
  • $\begingroup$ @Ruslan I differentiate your function and then set $g=1$. Result is $1/3$. $\endgroup$
    – Mick A
    Aug 3, 2014 at 14:50
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    $\begingroup$ The solution is nicely written up, and perfectly correct. $\endgroup$ Aug 3, 2014 at 14:51

4 Answers 4

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The answer is indeed $\frac13$. To check this, let's go another way: set $g=x-\frac2x$, as you did. Then express $x$ through $g$: $$x=\frac12\left(g\pm\sqrt{g^2+8}\right).$$

Then

$$f(g)=\sqrt{\frac12\left(g\pm\sqrt{g^2+8}\right)-1}.$$

Taking its derivative, we can find that it's real at $g=1$ for $\pm\to+$, and we get $$f'(1)=\frac13.$$

(The other solution with $\pm\to-$ is $-\frac{i}{6\sqrt2}$).

Why does your source say the answer is $1$? Maybe it's just a mistake of evaluating $f(1)$ instead of $f'(1)$, since $f(1)=1$.

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Differentiate both sides to get $f'(x-\frac{2}{x})(1+\frac{2}{x^2})=\frac{1}{2\sqrt{x-1}}$. Then use $x=2$ to evaluate, $f'(1)=\frac{1}{3}$.

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  • $\begingroup$ That's what he's done already. $\endgroup$
    – Ruslan
    Aug 4, 2014 at 6:02
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Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$

It is found that $x = 2$ or $x = -1$

Also $f(g(x))=f'(g(x)).g'(x)$ by the chain rule.

for $ g(x)=x-\frac {2}{x}$

$\Rightarrow g'(x) = 1+\frac {2}{x^2}$

for $f'(x)=\frac{1}{2} \frac {1}{\sqrt{(x-1)}}$

$f(g(x))=\left(1+\frac {2}{x^2}\right) \left(\frac {1}{2\sqrt{(x-1)}}\right)$

$\Rightarrow \frac {-5}{8}$ $\text{by differentiating and putting x=2 for already evaluated g(x) for x=1}\\$

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Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$

It is found that $x = 2$ or $x = -1$

Taking the positive value of $x$

Therefore $f(1) = \sqrt {x-1}$ and Thus $$f'(1) = \frac {1}{2}\cdot(x-1)^{\frac{1}{2}-1},$$ $ \text{1 for the value inside the function}$ $$\Rightarrow \frac{1}{2}\cdot(x-1)^{\frac{-1}{2}}$$ Putting the positive value $2$, we get $f'(1) = \frac{1}{2}$

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  • $\begingroup$ $f(1)=\sqrt{x-1}$ seems to be a nonsense. You've substituted $x=2$ on the left, but didn't on the right. As a result, you state that $\sqrt{x-1}=\text{const}$. $\endgroup$
    – Ruslan
    Aug 3, 2014 at 14:59
  • $\begingroup$ I feel I have substituted the value of the whole function and not the value of x on the left side, and it seems to be correct. Positive value of x is 2. $\endgroup$
    – Berry P J
    Aug 3, 2014 at 15:05
  • $\begingroup$ $f(1)$ usually means $f(x)|_{x=1}$. It's a constant. How can it depend on $x$? $\endgroup$
    – Ruslan
    Aug 3, 2014 at 15:10

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