Derivative of ln of arcsin of x/(x+1)
Find $\dfrac{dy}{dx}$ for: $$ y = \ln\left(\arcsin\frac{x}{x+1}\right). $$
Explanation of Method
This is a composition of three layers — a logarithm, an arcsine, and a rational function — so the chain rule must be applied twice, working from the outside in. Each layer contributes its own derivative rule: $\frac{d}{du}\ln u = \frac1u$ for the logarithm, $\frac{d}{dz}\arcsin z = \frac{1}{\sqrt{1-z^2}}$ for the arcsine, and the quotient rule for the innermost rational function $\frac{x}{x+1}$. Multiplying all three pieces together (and simplifying $1-z^2$ along the way) gives the final derivative.
- Apply the chain rule to the outer logarithm.
- Apply the chain rule to the arcsine, introducing $\frac{1}{\sqrt{1-z^2}}$.
- Differentiate the inner rational function $\frac{x}{x+1}$ with the quotient rule.
- Simplify $1-z^2$ and combine all the pieces into a single expression.
Step 1 · Differentiate the outer logarithm
Let $u(x) = \arcsin\dfrac{x}{x+1}$, so $y = \ln(u(x))$. By the chain rule:
$$ \frac{dy}{dx} = \frac{1}{u(x)}\cdot\frac{du}{dx} = \frac{1}{\arcsin\frac{x}{x+1}}\cdot\frac{du}{dx}. $$
Step 2 · Differentiate the arcsine
Let $z(x) = \dfrac{x}{x+1}$, so $u(x)=\arcsin(z(x))$. Since $\dfrac{d}{dz}\arcsin z = \dfrac{1}{\sqrt{1-z^2}}$, the chain rule gives:
$$ \frac{du}{dx} = \frac{1}{\sqrt{1-z^2}}\cdot\frac{dz}{dx}. $$
Step 3 · Differentiate the inner rational function
For $z(x) = \dfrac{x}{x+1}$, apply the quotient rule with $f(x)=x$, $g(x)=x+1$ (so $f'=1$, $g'=1$):
$$ \frac{dz}{dx} = \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2}. $$
Step 4 · Simplify $1-z^2$ and assemble the result
Compute $1-z^2$ using a common denominator:
$$ 1 - z^2 = 1 - \frac{x^2}{(x+1)^2} = \frac{(x+1)^2-x^2}{(x+1)^2} = \frac{2x+1}{(x+1)^2}, $$
using $(x+1)^2-x^2 = (x^2+2x+1)-x^2 = 2x+1$. Assuming $x+1>0$ (so that the square root simplifies without absolute values):
$$ \sqrt{1-z^2} = \frac{\sqrt{2x+1}}{x+1} \quad\Longrightarrow\quad \frac{1}{\sqrt{1-z^2}} = \frac{x+1}{\sqrt{2x+1}}. $$
Substituting into Step 2:
$$ \frac{du}{dx} = \frac{x+1}{\sqrt{2x+1}}\cdot\frac{1}{(x+1)^2} = \frac{1}{(x+1)\sqrt{2x+1}}. $$
Finally, substituting into Step 1:
$$ \frac{dy}{dx} = \frac{1}{\arcsin\frac{x}{x+1}}\cdot\frac{1}{(x+1)\sqrt{2x+1}} = \frac{1}{(x+1)\sqrt{2x+1}\,\arcsin\frac{x}{x+1}}. $$
Summary Table
| Layer | Rule applied | Contribution |
|---|---|---|
| $\ln(u)$ | Chain rule (log) | $\dfrac{1}{\arcsin\frac{x}{x+1}}$ |
| $\arcsin(z)$ | Chain rule (arcsine) | $\dfrac{x+1}{\sqrt{2x+1}}$ |
| $\dfrac{x}{x+1}$ | Quotient rule | $\dfrac{1}{(x+1)^2}$ |
Additional Notes
- The domain of $y$ requires $\dfrac{x}{x+1}\in[-1,1]$ (for the arcsine) and $\arcsin\dfrac{x}{x+1} > 0$ (for the logarithm to be defined), together with $2x+1\ge0$ from the simplification in Step 4 — in particular $x=0$ must be excluded, since $\arcsin(0)=0$ makes $\ln(0)$ undefined there.
- Simplifying $\sqrt{1-z^2}$ into $\dfrac{\sqrt{2x+1}}{x+1}$ before multiplying through is what collapses what looks like a three-story fraction into the clean final denominator $(x+1)\sqrt{2x+1}$.
- This problem is a good illustration of applying the chain rule "layer by layer" from the outside in — each of the three function layers (log, arcsine, rational) contributes exactly one factor to the final product.
Result
The derivative is: