From a Difference Quotient to a Slope
A derivative at a point is defined by a limit of nearby secant slopes. For the sine function, the point \(x=0\) gives a particularly direct example: the difference quotient simplifies to \(\sin h/h\). Evaluating its limit shows that the sine curve has slope \(1\) at the origin, provided angles are measured in radians.
The angle convention matters. The standard limit \(\lim_{h\to 0}\sin h/h=1\) is a radian limit. If the input is instead a numerical angle in degrees, the corresponding quotient has a different limit. We will derive the radian result, interpret it as a tangent slope, and then check what changes in degree measure.
The Derivative Definition at Zero
For a function \(f\), its derivative at \(x=a\), when the limit exists as a finite number, is
For each nonzero increment \(h\), the quotient is the slope of the secant line through the points on the graph with inputs \(a\) and \(a+h\). The derivative is the limiting value of those slopes as the second input approaches \(a\).
Since \(\sin 0=0\) and \(\sin(0+h)=\sin h\), the quotient for \(h\ne 0\) becomes
The expression is not defined at \(h=0\), because its denominator would be zero. That does not prevent the limit from existing: a limit examines values for nonzero \(h\) close to zero.
Worked Example: Simplifying the Sine Difference Quotient
Worked Example: Simplifying the Sine Difference Quotient
Use the derivative definition to write the difference quotient for \(f(x)=\sin x\) at \(x=0\), and simplify it for \(h\ne 0\).
Start with the definition:
Substitute \(f(x)=\sin x\):
Use \(0+h=h\) and \(\sin 0=0\):
This is the standard sine quotient. It is still a function of \(h\); the derivative requires its limit as \(h\) approaches zero, not its value at \(h=0\).
Check: Substituting the simplified expression back into the original quotient gives \((\sin h-0)/h=\sin h/h\), so the simplification preserves the quotient for every nonzero \(h\).
Why the Standard Sine Limit Equals One
The key trigonometric limit is
with \(h\) measured in radians. A geometric comparison explains why. First consider an angle \(h\) with \(0<h<\pi/2\) on the unit circle. The circle has radius \(1\). Compare the triangle inside the sector, the sector itself, and the larger triangle formed by extending one radius to the tangent line.
The inner triangle has area \(\frac12\sin h\). The sector has area \(\frac12 h\), because the area of a sector with radius \(1\) and radian angle \(h\) is \(\frac12h\). The outer triangle has area \(\frac12\tan h\). Their areas are ordered as inner triangle, sector, outer triangle. Therefore,
Multiplying all three parts by \(2\) gives \(\sin h<h<\tan h\). Since \(h>0\), division by \(h\) preserves the inequalities:
Using \(\tan h=\sin h/\cos h\), and \(\cos h>0\) for \(0<h<\pi/2\), the second inequality gives
Together, these inequalities show that
As \(h\) approaches zero from above, \(\cos h\) approaches \(\cos 0=1\). The squeeze theorem now gives \(\lim_{h\to0^+}\sin h/h=1\). For negative \(h\), set \(u=-h>0\). Since sine is odd, \(\sin(-u)=-\sin u\), and hence
As \(h\to0^-\), \(u=-h\to0^+\), so the quotient also approaches \(1\) from the negative side. Both one-sided limits agree; therefore the two-sided limit exists and equals \(1\).
Worked Example: Comparing Secant Slopes on Both Sides
Worked Example: Comparing Secant Slopes on Both Sides
Find the secant slopes from \((0,\sin 0)\) to the points with inputs \(h=\pi/6\) and \(h=-\pi/6\). Compare them with the limiting slope.
For \(h=\pi/6\), use the quotient \(\sin h/h\) and the exact value \(\sin(\pi/6)=1/2\):
The decimal is rounded to three places. For \(h=-\pi/6\), sine is odd, so \(\sin(-\pi/6)=-1/2\). Thus
The two secant slopes agree because both the numerator and denominator change sign for the negative increment. They are close to \(1\), but are not exactly \(1\) for these particular increments. As \(h\) approaches zero from either side, the standard limit shows that the secant slopes approach \(1\).
Check: In both calculations, multiplying the quotient by its denominator recovers the numerator: \((3/\pi)(\pi/6)=1/2\) on the positive side, and \((3/\pi)(-\pi/6)=-1/2\) on the negative side.
Applying the Limit to Find the Derivative
We have reduced the derivative definition to the standard limit, and the geometric argument established that limit in radians. Substitution into the definition gives
The graph of \(y=\sin x\) passes through \((0,0)\), since \(\sin 0=0\). Its derivative there is \(1\), so the tangent slope at the origin is \(1\). In geometric terms, near the origin the vertical change in the curve is approximately equal to the horizontal change when both are measured in the same radian-based coordinate system. This describes a local slope; it does not say that the sine curve is a straight line.
Worked Example: What Changes When the Input Is in Degrees?
Worked Example: What Changes When the Input Is in Degrees?
Suppose \(t\) is a numerical angle measured in degrees, and define \(g(t)\) to be the sine of that angle. Find the derivative at \(t=0\) from its difference quotient.
Convert the degree input \(t\) to radians: its radian measure is \(\pi t/180\). Thus \(g(t)=\sin(\pi t/180)\), and \(g(0)=0\). The quotient at zero is
Let \(u=\pi t/180\). Then \(t=180u/\pi\), so for \(t\ne0\),
As \(t\to0\), \(u\to0\). Applying the standard radian limit gives
The decimal is rounded to six places. The degree-based numerical input therefore gives a different derivative value from the radian-based input.
Check: The conversion \(u=\pi t/180\) has slope \(\pi/180\), and the quotient factors into that conversion factor times \(\sin u/u\). Since the latter tends to \(1\), the resulting limit is \(\pi/180\), as calculated.
Common Mistakes and Their Corrections
A common error is to substitute \(h=0\) directly into the quotient and conclude that the derivative is \(0/0\), so it does not exist. The incorrect calculation is
The correction is that the quotient is evaluated for \(h\ne0\), and then a limit is taken. The derivative is not defined by substituting zero into the quotient:
Another pitfall is using degree-mode calculator values while applying the radian limit. The limit \(\sin h/h=1\) as \(h\to0\) applies when \(h\) is in radians. If the input is a numerical number of degrees, the factor \(\pi/180\) must be included, as the degree example shows.
Why This Result Matters
This calculation is a model for finding a derivative directly from its definition. Begin with the function values at two nearby inputs, simplify the difference quotient, and then evaluate the limit. The trigonometric identity or limit used must match the units of the input; otherwise, the slope will be wrong.
The result also gives a precise meaning to the tangent slope at the origin. Individual secant slopes such as \(3/\pi\) are not the derivative, but their values approach the derivative as the second point moves toward the first. The limiting slope is \(1\), and the proof works from both the left and the right.
Check Your Understanding
Use the derivative definition and the radian convention when answering these questions.
- Write the difference quotient for \(f(x)=\sin x\) at \(x=0\), and simplify it for \(h\ne0\).
- State the standard limit used to evaluate the quotient, including the required angle unit.
- Why does the geometric comparison establish the limit for positive \(h\) close to zero?
- Explain why the quotient has the same limit from negative \(h\) as from positive \(h\).
- What is the tangent slope of \(y=\sin x\) at the origin, when \(x\) is measured in radians?
- Why does using a degree-valued input change the derivative value?
