Tangent Line
The "tangent line" is one of the most important applications of differentiation. The word "tangent" comes from the Latin word "tangere" which means "to touch". The tangent line touches the curve at a point on the curve. So to find the tangent line equation, we need to know the equation of the curve (which is given by a function) and the point at which the tangent is drawn.
Let us see how to find the slope and equation of the tangent line along with a few solved examples. Also, let us see the steps to find the equation of the tangent line of a parametric curve and a polar curve.
What is Tangent Line?
The tangent line of a curve at a given point is a line that just touches the curve (function) at that point. The tangent line in calculus may touch the curve at any other point(s) and it also may cross the graph at some other point(s) as well. The point at which the tangent is drawn is known as the "point of tangency". We can see the tangent of a circle drawn here.
If a line passes through two points of the curve but it doesn't touch the curve at either of the points then it is NOT a tangent line of the curve at each of the two points. In that case, the line is called a secant line. Here, we can see some examples of tangent lines and secant lines. The following shows a secant line PQ but which is NOT a tangent either at P or at Q.
Tangent Line Examples
Here is a typical example of a tangent line that touches the curve exactly at one point.
As we learned earlier, a tangent line can touch the curve at multiple points. Here is an example.
Again, the tangent line of a curve drawn at a point may cross the curve at some other point also. Here is the tangent line drawn at a point P but which crosses the curve at some other point Q.
The above line PQ can also be called the secant line. A secant line may also pass through any two points of the curve without the need to touch the curve at each of the two points.
Slope of Tangent Line
Let us consider a curve that is represented by a function f(x). Also, let us consider a secant line passing through two points of the curve P (x_{0}, f(x_{0})) and Q (x_{0} + h, f(x_{0} + h)). i.e., P and Q are at a distance of h units from each other.
Then the slope of the secant line using the slope formula is,
Slope of secant line = [f(x_{0} + h)  f(x_{0})] / (x_{0} + h  x_{0}) = [f(x_{0} + h)  f(x_{0})] / h
From the above figure, we can see that if Q comes very close to P (by making h → 0) and merges with P, then the secant line becomes the tangent line at P. i.e., the slope of the tangent line at P can be obtained by applying h → 0 to the slope of the secant line. So
Slope of Tangent at P = limₕ → _{0} [f(x_{0} + h)  f(x_{0})] / h
We know that this is nothing but the derivative of f(x) at x = x_{0} (by the limit definition of the derivative (or) first principles). i.e.,
 Slope of Tangent at P = f '(x_{0})
Therefore, the slope of the tangent is nothing but the derivative of the function at the point where it is drawn.
Slope of Tangent Line Formula
The slope of the tangent line of y = f(x) at a point (x_{0}, y_{0}) is (dy/dx)_{(x}_{0}, _{y}_{0}_{)} (or) (f '(x)) _{(x}_{0}, _{y}_{0}_{)}, where
 f'(x) is the derivative of the function f(x).
 (f '(x)) _{(x}_{0}, _{y}_{0}_{)} is the value obtained by substituting (x, y) = (x_{0}, y_{0}) in the derivative f '(x).
Note that we may have to use implicit differentiation to find the derivative f '(x) if the function is implicitly defined.
Tangent Line Equation
We know that the equation of a line with slope 'm' that is passing through a point (x_{0}, y_{0}) is found by using the pointslope form: y  y_{0} = m (x  x_{0}). Let us consider the tangent line drawn to a curve y = f(x) at a point (x_{0}, y_{0}). Then from the previous sections,
Slope of the tangent line, m = (f '(x)) _{(x}_{0}, _{y}_{0}_{)}
By substituting m, x_{0}, and y_{0} values in the pointslope form y  y_{0} = m (x  x_{0}) we can get the tangent line equation.
Thus, the tangent line formula is,
 y  y_{0} = (f '(x)) _{(x}_{0}, _{y}_{0}_{)} (x  x_{0})
Steps to Find the Tangent Line Equation
To find the tangent line equation of a curve y = f(x) drawn at a point (x_{0}, y_{0}) (or at x = x_{0}):
 Step  1: If the ycoordinate of the point is NOT given, i.e., if the question says the tangent is drawn at x = x_{0}, then find the ycoordinate by substituting it in the function y = f(x).
i.e., ycoordinate, y_{0} = f(x_{0}).  Step  2: Find the derivative of the function y = f(x) and represent it by f'(x).
 Step  3: Substitute the point (x_{0}, y_{0}) in the derivative f '(x) which gives the slope of the tangent (m).
 Step  4: Find the equation of the tangent using the pointslope form y  y_{0} = m (x  x_{0}).
Tangent Line Approximation
The concept of linear approximation just follows from the equation of the tangent line. i.e., The equation of the tangent line of a function y = f(x) at a point (x_{0}, y_{0}) can be used to approximate the value of the function at any point that is very close to (x_{0}, y_{0}). We can understand this from the example below.
Example of Tangent Line Approximation
Use the tangent line approximation to find the approximate value of ∛8.1.
Solution
We know that ∛8 = 2 and 8.1 very close to 8.
So we assume the function to be f(x) = ∛x and the point where the tangent is drawn to be x_{0} = 8.
Then (x_{0}, y_{0}) = (8, ∛8) = (8, 2).
The derivative of the function is f '(x) = (1/3) x^{2/3}
The slope of the tangent is, m = (f '(x))₍₈, ₂_{)} = (1/3) (8)^{2/3} = (1/3) (2^{3})^{2/3} = (1/3) (1/4) = 1/12
The equation of the tangent line is, y  y_{0} = m (x  x_{0})
y  2 = (1/12) (x  8)
y = x/12  2/3 + 2
y = x/12 + 4/3
Substituting y = f(x) here,
f(x) = x/12 + 4/3
Now, the approximate value of ∛8.1 can be obtained by substituting x = 8.1 here. Thus,
f(8.1) ≈ (8.1)/12 + 4/3
∛8.1 ≈ 2.008
We can check this with the calculator by finding the cube root of 8.1 and we can see it to be 2.008. This is how the tangent line approximation works.
Tangent Line Equation of Parametric Curve
Sometimes the function of the curve may not be given in the form y = f(x), instead, it might be represented in the parametric form. Let us see how to find the equation of a tangent line of a parametric curve in both 2D and 3D.
Tangent Line of Parametric Curve in 2D
If the curve in 2D is represented by the parametric equations x = x(t) and y = y(t), then the equation of the tangent line at t = a is found using the following steps:
 Find the point at which the tangent is drawn, (x_{0}, y_{0}) by substituting t = a in the given parametric equations.
i.e., (x_{0}, y_{0}) = (x(a), y(a)).  Find the derivative of the function using (dy/dt) / (dx/dt).
 Find the slope of the tangent (m) by substituting either t = a in the above derivative.
 Find the equation of the tangent line using y  y_{0} = m (x  x_{0}).
Tangent Line of Parametric Curve in 3D
Let the curve in 3D is defined by the parametric equations x = x(t), y = y(t), and z = z(t). Here are the steps to find the equation of the tangent line at a point t = t_{0}.
 Substitute t = a in each of the given equations to find the point (x_{0}, y_{0}, z_{0}) at which the tangent is drawn.
i.e., (x_{0}, y_{0}, z_{0}) = (x(t_{0}), y(t_{0}), and z(t_{0}))  Find the derivatives x'(t), y'(t), and z'(t).
 Substitute t = t_{0} in each of these derivatives to find the direction ratios <a, b, c> of the line.
i.e., <a, b, c> = <x'(t_{0}), y'(t_{0}), z'(t_{0})>  Find the equation of the tangent line using one of the following formulas:
x = x_{0} + at, y = y_{0} + bt, z = z_{0} + ct [OR]
(x  x_{0}) / a = (y  y_{0}) / b = (z  z_{0}) / c
We can see examples of these formulas in the "Examples" section below.
Tangent Line of Polar Curve
If the function is defined by polar equation r = r(t), then the equation of the tangent line at t = a is found by using the following steps:
 Find 'r' where r = r(a).
 Find the point (x_{0}, y_{0}) where the tangent is drawn using (x_{0}, y_{0}) = (r cos a, r sin a).
 Find dy/dx using the formula,\(\dfrac{d y}{d x}=\dfrac{\dfrac{d r}{d t} \sin (t)+r \cos (t)}{\dfrac{d r}{d t} \cos (t)r \sin (t)}\).
 Find the slope of the tangent using, m = (dy/dx)_{t = a}.
 Find the equation of the tangent line using y  y_{0} = m (x  x_{0}).
Important Notes on Tangent Line:
 The equation of tangent line of a curve y = f(x) at a point (x_{0}, y_{0}) is found using y  y_{0} = m (x  x_{0}),
where m = (f '(x))_{(x}_{0,} _{y}_{0}_{)}.  If θ is the angle made by the tangent line with the positive direction of the xaxis, then its slope is m = tan θ.
 Normal line and tangent line drawn for a curve at a point are perpendicular to each other and hence the slope of the normal = (1) / (slope of the tangent).
 A curve y = f(x) has horizontal tangents at the points where f '(x) = 0 as horizontal tangents are parallel to xaxis.
 A curve y = f(x) has vertical tangents at the points where f '(x) is undefined as horizontal tangents are parallel to yaxis.
 The tangent line equation is used to find the approximate values of the function in the neighborhood of the point at which the tangent is drawn.
☛ Related Topics:
Examples on Tangent Line Equation

Example 1: Find the equation of the tangent line of the curve y = 3x^{2}  4x at x = 1. Also, verify it.
Solution:
The point at which the tangent is drawn is, (x_{0}, y_{0}) = (1, f(1)) = (1, 3(1)^{2}  4(1)) = (1, 7).
The given curve is, f(x) = 3x^{2}  4x.
Its derivative is f'(x) = 6x  4.
The slope of the tangent is, m = f'(1) = 6(1)  4 = 10.
The equation of the tangent line is,
y  y_{0} = m (x  x_{0})
y  7 = 10 (x  (1))
y  7 = 10 (x + 1)
y  7 = 10x  10
y = 10x  3
Verification:
Let us draw the given function f(x) = 3x^{2}  4x and the tangent line graph of y = 10x  3 and verify whether it is a tangent.
The tangent line touches the given curve at a point and hence it is verified.
Answer: The tangent line equation is y = 10x  3.

Example 2: What is the tangent line equation of the curve x = cos t and y = sin t at t = π/2?
Solution:
The point at which the tangent is drawn is, (x_{0}, y_{0}) = (cos π/2, sin π/2) = (0, 1). (by using unit circle)
The curve is defined by parametric equations.
So dy/dx = (dy/dt) / (dx/dt) = (cos t) / ( sin t).
The slope of the tangent is, m = (dy/dx)ₜ ₌ π/₂ = (cos π/2) / (sin π/2) = 0/(1) = 0.
The equation of the tangent line is,
y  y_{0} = m (x  x_{0})
y  1 = 0 (x  0)
y  1 = 0
y = 1
Answer: The equation of the tangent line is y = 1.

Example 3: Find the parametric equations of the tangent line drawn to the curve given by the equations x = 2t^{2}, y = 3t, z = t^{3} at t = 2.
Solution:
The point at which the tangent line is drawn is,
(x_{0}, y_{0}, z_{0}) = (x(2), y(2), z(2)) = (2(2)^{2}, 3(2), 2^{3}) = (8, 6, 8).
Compute the derivatives
x'(t) = 4t, y'(t) = 3, z'(t) = 3t^{2}
The direction ratios of the tangent line are,
<a, b, c> = <x'(2), y'(2), z'(2)> = <4(2), 3, 3(2)^{2}> = <8, 3, 12>
The parametric equations of the tangent line are:
x = x_{0} + at, y = y_{0} + bt, z = z_{0} + ct
x = 8 + 8t, y = 6 + 3t, z = 8 + 12t
Answer:The parametric equations of the tangent line are x = 8 + 8t, y = 6 + 3t, z = 8 + 12t.
FAQs on Tangent Line
What is Tangent Line Definition?
The tangent line of a curve y = f(x) is a line that touches the curve at a point (x_{0}, y_{0}). Its slope (m) is found by substituting the point where it is drawn in the derivative f'(x) and its equation is found by using y  y_{0} = m (x  x_{0}).
How to Find the Slope of a Tangent Line?
The slope of a tangent line at a point is its derivative at that point. If a tangent line is drawn for a curve y = f(x) at a point (x_{0}, y_{0}), then its slope (m) is obtained by simply substituting the point in the derivative of the function. i.e., m = (f '(x))_{(x}_{0}, _{y}_{0}_{)}.
What is the Meaning of Point of Tangency?
A tangent line of a curve touches the curve at one point and that one point is known as the point of tangency. It is very important in finding the tangent line equation.
How to Find the Tangent Line Equation of y = f(x)?
To find the equation of tangent line of y = f(x) at x = x_{0}:
 Find the point (x_{0}, y_{0}) = (x_{0}, f(x_{0})).
 Find the slope using m by substituting (x, y) = (x_{0}, y_{0}) in f'(x), where f'(x) is the derivative of f(x).
 Find the tangent line equation using y  y_{0} = m (x  x_{0}).
How to Find the Tangent Line Equation of a Parametric Curve in 2D?
When the curve is defined by parametric equations x = x(t) and y = y(t), then the tangent line equation drawn at t = a is found using the steps below:
 Find the point (x_{0}, y_{0}) = (x(a), y(a)).
 Find the slope dy/dx using dy/dx = (dy/dt) / (dx/dt).
 Find the slope of the tangent using m = (dy/dx)_{t = a}
 Find the equation of the tangent line using y  y_{0} = m (x  x_{0}).
Can Tangent Line Cross the Curve?
The only condition for a line to be a tangent of a curve at a point is that the line should touch the curve at that point. But it can cross the graph at any other point(s).
How to Find the Tangent Line Equation of a Polar Curve?
If a curve is defined by the polar equation r = r(t), then the tangent line equation drawn at t = a is found using the steps below:
 Find r = r(a).
 Find the point (x_{0}, y_{0}) = (r cos a, r sin a).
 Find the slope dy/dx using dy/dx = \(\dfrac{\dfrac{d r}{d t} \sin (t)+r \cos (t)}{\dfrac{d r}{d t} \cos (t)r \sin (t)}\)
 Find the slope of the tangent using m = (dy/dx)_{t = a}.
 Find the equation of the tangent line using the pointslope form, y  y_{0} = m (x  x_{0}).
How to Find Where the Tangent Line is Vertical?
The slope of a vertical line is undefined. Also, we know that the slope of a tangent line is equal to the derivative. Thus, to see where the tangent line is vertical, just see where the derivative is undefined. If the derivative is rational, just set the denominator = 0 and solve.
How to Find the Tangent Line Equation of a Parametric Curve in 3D?
When a curve in three dimensions is defined by the parametric equations x = x(t), y = y(t), and z = z(t), then the tangent line equation drawn to it at x = t_{0} is found as follows:
 Find the point using (x_{0}, y_{0}, z_{0}) = (x(t_{0}), y(t_{0}), and z(t_{0}))
 Find the derivatives of given functions. i.e., find x'(t), y'(t), and z'(t)
 Find the direction of the line using <a, b, c> = <x'(t_{0}), y'(t_{0}), z'(t_{0})>
 Find the tangent line equation using x = x_{0} + at, y = y_{0} + bt, z = z_{0} + ct (parametric form) or (x  x_{0}) / a = (y  y_{0}) / b = (z  z_{0}) / c (cartesian form)
How to Find Horizontal Tangent Line Equation?
A horizontal tangent is parallel to xaxis and hence its slope is zero. We know that the slope is nothing but the derivative of the function. So to find the points where there are horizontal tangents just set the derivative of the function to zero and solve. After getting the points, we can find the equation of the horizontal tangent line using the pointslope form.
How to Find Vertical Tangent Line Equation?
A vertical tangent is parallel to yaxis and hence its slope is undefined. As the slope is nothing but the derivative of the function, to find the points where there are vertical tangents, see where the derivative of the function becomes undefined (probably set the denominator of the derivative to zero to find it). After getting the points, we can find the equation of the vertical tangent line using the pointslope form.
What is the Difference Between Vertical and Horizontal Tangent Lines?
The slope of a horizontal tangent line is 0 (i.e., the derivative is 0) as it is parallel to xaxis. The slope of a vertical tangent line is undefined (the denominator of the derivative is 0) as it is parallel to the yaxis.
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