= Solution
For a nonzero <isogeny of elliptic curves> $\phi:E_1\to E_2$, its <degree of an isogeny> is the degree of the induced extension of <function fields>. Since $\phi$ commutes with negation, its first coordinate is a <rational function> of $x_1$ alone. Write it in lowest terms as
$$
x_2(\phi(P))=R(x_1(P))=\frac{A(x_1(P))}{B(x_1(P))},\qquad\gcd(A,B)=1.
$$
The coordinate maps $x_i:E_i\to\mathbb P^1$ are <finite morphisms> of degree two. Multiplying <function field> degrees in the commutative diagram $x_2\circ\phi=R\circ x_1$ gives the <degree of an isogeny from its x-coordinate map>:
$$
\boxed{\deg\phi=\max\{\deg A,\deg B\}.}
$$
This is the total degree, so the argument also covers an inseparable <isogeny of elliptic curves>. Counting distinct points in the <kernel of an isogeny> would give only its separable degree. The zero <group homomorphism> is assigned degree zero separately.
On a <Short Weierstrass form> $y^2=x^3+ax+b$ in <characteristic of a field> other than two, the <elliptic-curve addition formula> gives
$$
x(2P)=\left(\frac{3x^2+a}{2y}\right)^2-2x
=\frac{x^4-2ax^2-8bx+a^2}{4(x^3+ax+b)}.
$$
There is no cancellation: at a root $r$ of $f(x)=x^3+ax+b$, the numerator equals $f'(r)^2$, which is nonzero because the <elliptic-curve discriminant> is nonzero. The numerator has degree four and the denominator degree three. Hence
$$
\boxed{\deg[2]=4.}
$$
For completeness, the same answer holds in characteristic two, where the displayed short equation is not a nonsingular model. On a general <Weierstrass equation of an elliptic curve> use
$$
x(2P)=\frac{x^4-b_4x^2-2b_6x-b_8}{4x^3+b_2x^2+2b_4x+b_6},
$$
where $b_2=a_1^2+4a_2$, $b_4=a_1a_3+2a_4$, $b_6=a_3^2+4a_6$, and $b_8=a_1^2a_6+4a_2a_6-a_1a_3a_4+a_2a_3^2-a_4^2$. The <resultant> of this numerator and denominator is $\Delta^2$. Their resultant is therefore nonzero in every nonsingular case; even when the denominator drops degree, the numerator retains degree four.
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