If
\[
y=\tan(3\tan^{-1}x),
\]
then
\[
(1-3x^2)\frac{d^2y}{dx^2}-12x\frac{dy}{dx}
\]
is equal to:
Show Hint
For expressions involving \(\tan(3\tan^{-1}x)\), first use the identity
\[
\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.
\]
This makes differentiation easier.
Step 1: Simplify y using the triple angle formula.
\(y=\tan(3\tan^{-1}x)=\frac{3x-x^3}{1-3x^2}\). So \(y(1-3x^2)=3x-x^3\).
Step 2: Differentiate once.
Differentiate: \(y'(1-3x^2)-6xy=3-3x^2=3(1-x^2)\). So \(y'(1-3x^2)=3(1-x^2)+6xy\).
Step 3: Differentiate again to get the expression.
Differentiate \(y'(1-3x^2)=3(1-x^2)+6xy\): \(y''(1-3x^2)-6xy'=-6x+6y+6xy'\Rightarrow(1-3x^2)y''-12xy'=6(y-x)\). \[\boxed{6(y-x)}\]