Derivative of arcsin(x^2): Chain Rule Proof, Mistakes

Substituting the inner function into the formula

The base rule is $\frac{d}{du}\arcsin u = \frac{1}{\sqrt{1-u^{2}}}$. With $u = x^{2}$, every $u$ in that formula becomes $x^{2}$, including the one being squared inside the radical, and then the inner derivative $u' = 2x$ multiplies the result.

The step that decides the whole problem is $\left(x^{2}\right)^{2} = x^{4}$. Exponents multiply under a power of a power, so the radicand is $1-x^{4}$. Getting this wrong changes the domain, the graph, and every value the derivative takes.

Domain and the vertical tangents at the ends

Arcsine accepts inputs between $-1$ and $1$, so $\arcsin\left(x^{2}\right)$ needs $x^{2}\le 1$, giving $-1\le x\le 1$. Since $x^{2}\ge 0$, the outputs only ever run from $0$ up to $\frac{\pi}{2}$.

At $x = \pm 1$ the radicand $1-x^{4}$ is $0$, so the derivative is undefined and the graph has vertical tangents at both endpoints. At values like $x = 0.2, 0.5, 0.8$ the derivative is finite and rising ($0.400$, $1.033$, $2.082$), steepening as $x$ approaches $1$.

The factor $2x$ also makes the derivative negative for $x<0$ and positive for $x>0$, so the function dips to a minimum of $0$ at the origin. That matches $\arcsin\left(x^{2}\right)$ being an even function.

The mistakes students make

Graders see the same three answers on this one every year, and all three come from mishandling the inner function $x^{2}$.

• Giving $\frac{2x}{\sqrt{1-x^{2}}}$. The inner function must be substituted inside the radical too, which turns $1-u^{2}$ into $1-x^{4}$.
• Losing the factor $2x$ and handing in $\frac{1}{\sqrt{1-x^{4}}}$, which handles the substitution correctly but drops the inner derivative the chain rule requires.
• Treating $\arcsin\left(x^{2}\right)$ as $\left(\arcsin x\right)^{2}$. That different function has derivative $\frac{2\arcsin x}{\sqrt{1-x^{2}}}$, and the two are not interchangeable.