How do you differentiate $f(x)=x/ln(sqrt(1/x))$ using the chain rule?

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Answer 1 · 2016-02-10T14:37:01

$f'(x) = (-2 ln x + 2)/(ln x)^2$

1) Simplifying

First of all, you can simplify using the following logarithmic laws:

[1] $" "log_a(m/n) = log_a(m) - log_a(n)$

[2] $" "log_a(m^r) = r * log_a(m)$

[3] $" "b^(1/2) = sqrt(b)$

Thus, you have

$ln(sqrt(1/x)) stackrel("[3] ")(=) ln [(1/x)^(1/2)] stackrel("[2] ")(=) 1/2 ln(1/x) stackrel("[1] ")(=) 1/2 (ln(1) - ln(x))$

Now, $log_a(1) = 0$ always holds for any basis $a > 0$ , $a != 1$ .

Using this, you have,

$ln(sqrt(1/x)) = 1/2 (ln(1) - ln(x)) = - 1/2 ln(x)$

Now your function can be simplified to:

$f(x) = x / ln(sqrt(1/x)) = (-2x) / ln(x)$

2) Differentiating

It doesn't make sense to apply the chain rule to differentiate $f(x)$ in this form. However, you can use the quotient rule:

If $f(x) = (g(x)) / (h(x))$ , then

$f'(x) = (g'(x) h(x) - g(x) h'(x))/(h^2(x))$

In your case, the derivatives of $g(x)$ and $h(x)$ are

$g(x) = -2x " " =>" " g'(x) = -2$
$h(x) = ln(x) " " =>" " h'(x) = 1/x$

Thus, your derivative is:

$f'(x) = (-2 * ln x + 2x * 1/x) / (ln x)^2 = (-2 ln x + 2)/(ln x)^2$

How do you differentiate f(x)=x/ln(sqrt(1/x))f(x)=x/ln(sqrt(1/x)) using the chain rule?