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The directional derivative of a scalar function of two variables at a point represents the rate of change of the function's output when the input is nudged infinitesimally in a chosen direction. Geometrically, it is the slope of the tangent line to the curve formed by intersecting the surface with a vertical plane passing through the point and parallel to the direction vector.

Conditions: The function is a scalar-valued function of two variables.; The direction is specified by a vector in the input plane.; The step size approaches zero.

The directional derivative of a scalar function of two variables at a point represents the rate of change of the function's output when the input is nudged infinitesimally in a chosen direction. Geometrically, it is the slope of the tangent line to the curve formed by intersecting the surface with a vertical plane passing through the point and parallel to the direction vector.

Conditions: The function is a scalar-valued function of two variables.; The direction is specified by a vector in the input plane.; The step size approaches zero.

The limit definition fails to produce a finite derivative for $f(x)=\sqrt{x}$ at $x=0$. While the formula yields $\frac{1}{2\sqrt{x}}$ for $x>0$, substituting $x=0$ results in division by zero, indicating that the slope of the tangent line is vertical and the derivative is undefined (infinite) at that point.

Conditions: $f(x)=\sqrt{x}$.; The point of evaluation is $x=0$.; The limit is taken as $h \to 0^+$ (since the domain is $x \ge 0$).

The limit definition fails to produce a finite derivative for $f(x)=\sqrt{x}$ at $x=0$. While the formula yields $\frac{1}{2\sqrt{x}}$ for $x>0$, substituting $x=0$ results in division by zero, indicating that the slope of the tangent line is vertical and the derivative is undefined (infinite) at that point.

Conditions: $f(x)=\sqrt{x}$.; The point of evaluation is $x=0$.; The limit is taken as $h \to 0^+$ (since the domain is $x \ge 0$).

To find the derivative of $f(x)=\sqrt{x}$, substitute into the difference quotient $\frac{\sqrt{x+h}-\sqrt{x}}{h}$. Multiply the numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt{x}$ to rationalize the numerator.

Conditions: $f(x)=\sqrt{x}$.; $x > 0$ for the derivative to be finite.; $h$ is nonzero during the algebraic manipulation.

To find the derivative of $f(x)=\sqrt{x}$, substitute into the difference quotient $\frac{\sqrt{x+h}-\sqrt{x}}{h}$. Multiply the numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt{x}$ to rationalize the numerator.

Conditions: $f(x)=\sqrt{x}$.; $x > 0$ for the derivative to be finite.; $h$ is nonzero during the algebraic manipulation.