Calculus distance maximizing 
Point $(x, y)$ satisfies the inequality $x^4 + y^2 \leqslant 1$. Determine the largest possible distance from the origin for $(x, y)$.

So the largest distance will result when $x^4+x^2=1$.
The distance from the origin for point $(x,y)$ is $d=\sqrt{x^2+y^2}$, but from the given inequality we see that $y^2=1-x^4$ therefore we get $d(x)=\sqrt{x^2+1-x^4}$. I know that in order to maximize distance we should find the derivative of $d(x)$, but it seems rather peculiar to differentiate that expression. Is there another way I could go with this?

Attempt:
Constraint: $x^4+y^2 \le 1$;  $y^2\le 1-x^4;$
Maximize : $x^2+y^2\le x^2+1-x^4=$
$-(x^4-x^2)+1=$
$-[(x^2-1/2)^2-1/4]+1=-(x^2-1/2)^2+5/4\le 5/4$.
Maximum attained at $x^2=1/2$; 
$y^2=5/4 -1/2=3/4.$
