The point of tangency is .
The slope of the tangent line is .
The equation of the tangent line is .
In this section we learn the definition of the derivative and we use it to solve the tangent line problem.
The idea of a tangent line is that if we zoom in on a point on the graph of a smooth function, , then the graph looks like a straight line. This line is the tangent line and the point is called the point of tangency.
The interactive graph below shows a function and its tangent line at a point. You can drag the point along the curve and see how the tangent line changes. You can also zoom in on the curve to see that the tangent line approximates the curve locally. If you zoom in far enough, the curve and the tangent line become nearly indistinguishable.
The tangent line problem is to find the equation of the tangent line to the graph of at the point . From the point-slope form of the equation of a line, the equation of the tangent line has the form where is the slope of the line. The derivative, denoted and read ‘f prime of x’, is the mathematical function which will give us the slope of the tangent line at any point. In particular, the slope of the tangent line at the point is denoted by . Thus, using , the equation of the tangent line can be written as
We will find the second form useful later when we study linear approximation. By distributing and adding, we can convert the point-slope form into the more common slope-intercept form, . For example, if the point is and , then the point-slope form of the line is Distributing the 3 and then adding 2, gives the slope intercept form:
Our objective is to discover a method for computing the derivative, , of a given function, . Since the derivative represents the slope of the tangent line to the graph of a function at a point, we begin by recalling the formula for the slope of a line between two points, and :
To find the slope of the tangent line to the graph of at the general point , we first compute the slope of the secant line between the point and a “nearby” point : We can simplify the denominator: This quantity is known as the difference quotient.
The slope of the tangent line, is obtained by letting which has the effect of moving the point towards the point .
This gives us the definition of the derivative:
Below is an interactive graph that shows both a tangent line (in red) and a secant line (in green). Consider the red dot to be the point . Move the red dot to a point of your choosing. The green dot represents the point . Move the green dot toward the red dot. Observe that when the green dot is very close to the red dot (i.e. is very small), the secant line becomes indistinguishable from the tangent line. If the green dot is to the right of the red dot, then is positive, and if the green dot is to the left of the red dot, then is negative.
We now compute the derivative of several different functions.
Thus the derivative of is and this tells use the slope of the tangent line at a given -value. The process of obtaining the derivative is called differentiation, and the mathematical symbol for the differentiation operator is so that we can write: This is essentially the notation used by German polymath, philosopher and co-discoverer of calculus, Gottfried Wilhelm Leibniz circa 1700. Now let’s use the derivative to discuss tangent lines and their slopes.
At the point on the graph of the parabola, , the slope of the tangent line is given by
That is, we plug the x-coordinate, , into the function .
The points and are also on the parabola. At the point , the slope of the tangent line
is and at the point the slope is
The graph below shows the parabola, , along with the tangent lines at and .
The point of tangency is .
The slope of the tangent line is .
The equation of the tangent line is .
Step 2. Compute the numerator of the difference quotient, .
Using the result of step 1, we have
Step 3. Divide by to obtain the difference quotient.
Using the result of step 2,
Step 4. Compute the limit of the difference quotient.
Using step 3, we have
Thus, the derivative of is . This can also be written using Leibniz notation as
In general, if y = f(x) then the notations are all valid ways to express the derivative. Furthermore, the following are valid ways to express the derivative evaluated at a point , as when finding the slope of a tangent line:
Step 1. Compute :
Step 2. Compute the numerator of the difference quotient:
Step 3. Divide by to obtain the difference quotient:
Step 4. Take the limit as . The derivative is
Step 1. Compute :
Step 2. Compute the numerator of the difference quotient:
Step 3. Divide by to obtain the difference quotient:
Step 4. Take the limit as . The derivative is
We have \begin{align*} f(x+h) &= (x+h)^3 \\ &= (x+h)(x+h)^2 \\ &= (x+h)(x^2 + 2xh + h^2)\\ &= x^3 + 2x^2h + xh^2 + x^2h + 2xh^2 + h^3\\ &= x^3 + 3x^2h + 3xh^2 + h^3. \end{align*}
Now we use the definition: \begin{align*} f'(x) &= \lim _{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim _{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3)- x^3}{h}\\[5pt] &= \lim _{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}\\[5pt] &= \lim _{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}\\[5pt] &= \lim _{h \to 0} (3x^2 + 3xh + h^2) \\ &= 3x^2. \end{align*}
To recap, Try to notice a pattern in the form of the derivatives of and .
To recap, which in the notation of exponents, where , we can write
Now for the derivative: \begin{align*} f'(x) &= \lim _{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim _{h \to 0} \frac{\sqrt{2x+2h+1}- \sqrt{2x+1}}{h}\\[5pt] &= \lim _{h \to 0} \frac{\sqrt{2x+2h+1}- \sqrt{2x+1}}{h} \cdot \frac{\sqrt{2x+2h+1}+ \sqrt{2x+1}}{\sqrt{2x+2h+1}+ \sqrt{2x+1}} \\[5pt] &= \lim _{h \to 0} \frac{(2x+2h+1) - (2x+1)}{h(\sqrt{2x+2h+1}+ \sqrt{2x+1})}\\[5pt] &= \lim _{h \to 0} \frac{2h}{h\sqrt{2x+2h+1}+ \sqrt{2x+1}}\\[5pt] &= \lim _{h \to 0} \frac{2}{\sqrt{2x+2h+1}+ \sqrt{2x+1}}\\[5pt] &= \frac{2}{\sqrt{2x+1}+ \sqrt{2x+1}}\\[5pt] &= \frac{2}{2\sqrt{2x+1}}\\[5pt] &= \displaystyle{\frac{1}{\sqrt{2x+1}}}. \end{align*}
In the notation of Leibniz,
Step 1. Compute :
Step 2. Compute the numerator of the difference quotient, :
Step 3. Divide by to obtain the difference quotient: (multiply by the conjugate radical and cancel )
Step 4. Take the limit as . The derivative is
Using the differential operator, , we can write:
\begin{align*} f'(x) &= \lim _{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim _{h \to 0} \frac{\frac{2}{x+h +3}- \frac{2}{x+3}}{h}\\[5pt] &= \lim _{h \to 0} \frac{\frac{2(x+3) - 2(x+h+3)}{(x+h+3)(x+3)}}{h} \\[5pt] &= \lim _{h \to 0} \frac{\frac{2x+6 - 2x-2h-6}{(x+h+3)(x+3)}}{h} \\[5pt] &= \lim _{h \to 0} \frac{-2h}{(x+h+3)(x+3)}\cdot \frac{1}{h}\\[5pt] &= \lim _{h \to 0} \frac{-2}{(x+h+3)(x+3)} \\[5pt] &= -\frac{2}{(x+3)^2}. \end{align*}
To recap, the derivative of is .