MATH 136
Geometric Interpretation of Derivative:
Slope of the Tangent Line
Given the graph of a function y = f (x) , we can find the equation of the line tangent to
the graph at a specific point x = x0 . To find the equation of this “tangent line,” we need
a point (x0 , y0 ) on the line and the slope m of the line. The point/slope equation of the
line is then y − y 0 = m (x − x 0 ) , or y = m (x − x0 ) + y0 .
Because the point (x0 , y0 ) €is also on the graph of y = f (x) , the point will be
(x0 , f (x 0 )) . The slope is given by the derivative f ′(x0 ) at point x0 . This geometric
interpretation of the derivative is stated below:
Given the graph of a function y = f (x) , the derivative f ′(x0 ) at a specific
point x0 gives the slope of the tangent line to the graph of y = f (x) at x0 .
The derivative f ′(x) is a function of x . As x varies, f ′(x) also varies. As x varies,
there will be different tangent lines, thus there will be different values of the slopes
f ′(x) of these tangent lines.
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The equation of the tangent line to the graph of
y = f (x) at x = x0 is given by
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y = f ′(x0 )(x − x 0 ) + f (x0 ) .
Example 1. Let f (x) = x 2 − 4x − 4 . Find the equation of the tangent line to the graph of
f at x = 3 . (This graph is illustrated above.)
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Solution. At x = 3, the point on the graph is (3, f (3)) = (3, − 7) . The derivative of f is
f ′(x) = 2x − 4 ; thus, the slope of the tangent line at x = 3 is f ′(3) = 2 .
The equation of the tangent line at x = 3 is then
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y = 2(x − 3 ) €
+ (−7) = 2x −13.
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Undefined Tangents: Endpoints, Vertical Tangents, Sharp Points
Consider the graphs of y1 = 16 x − 1 − 9(x − 5)1/ 3 and y2 = 20(x + 2)2 /5 .
At endpoints and at “sharp” points, there is not a “well-defined” tangent line; thus,
the derivative at such points (i.e. the slope of the tangent line at these points) will not
exist. Also, when a graph becomes nearly vertical at point, then the slope of the tangent
line approaches ± ∞. Again, the derivative at this point will not exist.
Example 2. Let y1 = 16 x − 1 − 9(x − 5)1/ 3 , for x ≥ 1 (the top graph above). At what
points is the tangent line undefined? That is, where is the derivative undefined?
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Solution. Because of the square root, the
function is only defined when x −1 ≥ 0. That is,
the domain is x ≥ 1. So there is an endpoint at x = 1. The derivative of y1 is
dy1
1
1
−1/2
= 16 × (x − 1)
− 9 × (x − 5) −2/ 3 =
dx
2
€ 3
€
8
3
.
€−
x − 1 (x − 5)2 / 3
We can see algebraically that the derivative is undefined when the denominator
terms are 0, which is at x = 1 and at x = 5 . There is not a well-defined tangent line at the
endpoint x = 1. At x = 5 , the graph is becoming vertical; thus, the slope of the tangent
line at x = 5 is infinite. In this case,
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lim
x → 5−
8 €
3
= lim
−
x − 1 (x − 5)2/ 3
x → 5+
8
3
= –∞.
−
x − 1 ( x − 5)2/ 3
8
3
−
2/ 3 = +∞. As x decreases to 1, the slope of the
x →1+ x − 1 (x − 5)
tangent line is increasing to +∞.
Note that
lim
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Example 3. Let y2 = 20(x + 2)2 /5 (the second graph above). At what points is the
tangent line undefined?
8
dy 2
2
−3/ 5
= 20 × (x + 2)
=
. The derivative is
dx
5
(x + 2)3 / 5
undefined at x = −2 ; thus, there is not a well-defined tangent line at the “sharp” point
when x = −2 . In this case,
Solution. The derivative of y2 is
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8
8
= –∞ and lim
= +∞.
3/
5
3/ 5
x → −2 − (x + 2)
x → −2 + (x + 2)
lim
As x increases to –2, the slope of the tangent line is decreasing to –∞; but as x decreases
to –2, the slope of the tangent line is increasing to +∞.
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