How do you find the slope of a curve at a given point?

Category: Derivatives »

Course: Calculus »

Relevant MM Lessons:This question involves a few major concepts across several lessons - for a deeper dive and for a full gambit of practice and common tricks, make sure to check out each Relevant Mister Math Lesson in the list above.

Tangent Slopes

One of the major purposes of Calculus is to find the slope of a curvy function at a specific point. Since the function is indeed curvy, and the slope is ever-changing, we often refer to the instantaneous slope at a given $x$ value as the tangent slope of the curve at that point.Ultimately, we cannot simply "eyeball" the answer. We have to use Calculus. The tangent slope of a function at a specified point is equal to the value of the function's derivative function at that $x$ value. Depending on the instructions we are given, we can get this via two approaches.

The Easy Way

If your instructions don't specify a method, you will of course choose the easier of the two ways. This method has two major steps, both of which are typically fast calculations:
  1. For your given $f(x)$, find $f'(x)$, which is the derivative function of $f(x)$.
  2. Evaluate $f'(x)$ at the $x$ value that you are interested in. In other words, if you want the derivative of $f$ at $x=a$, evaluate $f'(a)$.
Finding $f'(x)$ depends entirely on the function you're looking at. Polynomial Derivatives » are done using the "Power Rule". Exponential or Log » terms require their own formulas, as do trigonometric functions ». However, once you get $f'(x)$, just plug in $x=a$ to get the slope at $x=a$. Quick examples include:The slope of $f(x) = 2x^2-3x+1$ at $x=-2$:$$\blacktriangleright \,\, f'(x) = 4x - 3$$$$f'(-2) = -8 - 3 = -11$$The slope of $g(x) = 4\ln(x) + \frac{1}{x^2}$ at $x=1$:$$\blacktriangleright \,\, g'(x) = \frac{4}{x} - \frac{2}{x^3}$$$$g'(1) = \frac{4}{1} - \frac{2}{1} = 2$$The slope of $w(x) = \sin(x) + e^{-x}$ at $x = 0$:$$\blacktriangleright \,\, w'(x) = \cos(x) - e^{-x}$$$$w'(0) = \cos(0) - e^{0} = 1 - 1 = 0$$

The Long Way

If (and really, only if) your instructions require you to find the tangent slope using the "limit definition of the derivative" or just "the definition of the derivative" or the "limit of the difference quotient", you are being asked to do something a little more specific. It yields the same answer as the "Easy Way" above, but the method and calculation is the long form limit definition of how $f'(a)$ is rigorously defined.The major steps in this case are
  1. Write down $f(x)$.
  2. Under that, write down and simplify $f(x+h)$.
  3. Plug both $f(x)$ and $f(x+h)$ into the limit definition of the derivative formula, and evaluate the limit by simplifying the fraction.
The limit definition of the derivative uses the formula$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$$Here's one quick example.Find the slope of $f(x)= 2x^2-3x+1$ at $x=-2$ using the limit definition of the derivative.$\blacktriangleright$ We just did this one above, so we know what answer to expect.$$\begin{align}{rl} f(x) & = 2x^2-3x+1 \\ \therefore f(-2) & = 15 \\ f(x+h) & = 2(x+h)^2 - 3(x+h) + 1 \\ \therefore f(-2 + h) & = 2h^2 - 11h + 15 \end{align}$$$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$$$$=\lim_{h \to 0} \frac{[2h^2 - 11h + 15] - [15]}{h}$$$$=\lim_{h \to 0} \frac{2h^2 - 11h}{h}$$$$=\lim_{h \to 0} \frac{h(2h-11)}{h}$$$$=\lim_{h \to 0} 2h - 11 = -11$$The same answer we got with the "fast" way!