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## AP®︎ Calculus BC (2017 edition)

### Course: AP®︎ Calculus BC (2017 edition)>Unit 3

Lesson 4: Derivatives of negative and fractional powers with power rule

# Worked example: Tangent to the graph of 1/x

A practice problem for finding the y-intercept of a tangent line to f(x) = 1/x. Created by Sal Khan.

## Want to join the conversation?

• Why is y = 1/k and x = k ?

How is that possible ?
• That's how the function is defined. At the beginning of the video Sal said we are dealing with the function y = 1/x. The specific question we are answering is what is the y-intercept when x = k. When x = k, y is equal to 1/k.
• At , why is the derivative of the curve the same as the slope of its tangent line?
• Because the derivative was defined to be exactly that.
• What is Local linearization?
• The tangent line at a point on a curve can approximate points close to that value.
• what about tangent lines using a circle ?
(1 vote)
• A circle is not a function, and so isn't applicable here. A function must have no more than one y value for any given x value. To understand the concept of a function better, watch the videos in this section: http://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-relationships-functions/cc-8th-function-intro/v/testing-if-a-relationship-is-a-function.

However, you can find the tanget line of a circle by using "implicit" differentiation and providing a specific point on the circle. Think about a circle for a minute: for most of the x values on a circle, there are two corresponding y values, one above the center point and one below the center point. You can't determine the tanget line of a circle with just an x value because it matters which y value corresponding to the x value you're choosing. With an ellipse or circle, the slope is determined by BOTH the x and the y variable (unlike a function, whose slope is defined in terms of x only). Sal has so good videos on the topic here: http://www.khanacademy.org/math/calculus/differential-calculus/implicit_differentiation/e/implicit-differentiation.

That said, once you find the derivative of a circle through implicit differentiation, finding the y-intercept of a line tanget to the circle, given an arbitrary k where x = k, works just the same as it does in this video.
• This is kind of the same thing as what he does in this video but it is sort of backwards and I'm having trouble with it. If you have the equation of a curve and are asked to find the points where the tangent line has a y-intercept of 5, how do you go about that?
• Let us consider the facts we know about tangent lines:
1. They must intersect the function at the point of tangency.
2. They must have the same slope as does the function at the point of tangency.

Using the point-slope form of the line, this means:
y-y₁ = m(x-x₁)
Where (x₁, y₁) is the point of tangency.
Now, we need to replace m with the first derivative of the curve, since the first derivative can be interpreted as the slope of the curve (as a function instead of a constant).
y-y₁ = (f')(x-x₁)
where f' is the first derivative of the function.
We are now given an additional piece of information, the y=intercept of the tangent line. in other words, the tangent line goes through the point (0, 5). We plug that in for x and y:

5-y₁ = (f')(-x₁)

Note: we would use the original function (not its derivative) to give us enough information to solve for two unknowns. We plug in the point of tangency (x₁,y₁) into the original function. This would give us two equations and two unknowns, which would allow us to solve for (x₁,y₁) (there may be multiple solutions).
• The slope of a line can be written as tan = y/x. if at k f(x) is 1/k then tan = (1/k)/k = 1/(k^2) but when derivative is taken it gives us -1/(k^2) why is that?
• Notice that the slope of the line you are finding gives the slope from the origin to (k, f(k)) for some some x = k. But the derivative of f(x) at x=k gives the slope exactly at x =k. These slope from the origin is a secant line, and the second is an instantaneous slope. Thus, their values are not the same in this example.
• Are there more examples? I don't understand what he did to make x to the -1 turn into a -1 times x to the -2
(1 vote)
• When using the power rule you should multiply the x coefficient (1) by the exponent (-1), and then reduce the exponent by 1.
• at 2.40, why is the exponent of -1 in f(x) became -2 in f ' (x)?
(1 vote)
• The equation is f(x)=x^-1
Since the power rule states that
d/dx (x^n) = nx^(n-1)
Thus, f'(x) = -1x^(-1-1) = -1x^-2
• I have a silly question, In the derivative introduction videos Sal said that for a function to be differentiable it must be continuous, Defined, has no vertical asymptotes and the limit of the derivative must exist if there are any other rules please tell me
my question is how can a function like 1/x not be continuous and has a vertical asymptote still has a derivative of -x^-2 ?
(1 vote)
• Why is the y-value 1/k ?
(1 vote)
• Remember that the function is defined as f(x) = 1/x, where x does not equal 0. So if f(x) = 1/x, then any number we plug into the function is defined as 1 over that number.

So if we said that x = 1, then f(1) = 1/1. If x = 2, then f(2) = 1/2. If x = pi, then f(pi) = 1/pi. And so on for any number that is not zero. In this example, x = k (k is an arbitrary constant), and therefore f(k) = 1/k.

Now, keep in mind that f(x) is just a fancy way of saying y. We can define our x,y ordered pairs in terms of (x,y). Or, we can define them in terms of (x,f(x)). It's the same thing. But here, we know from the problem that x = k (note that Sal double-underlined that fact). So we can substitue k for x, which yields (k,f(k)). Since we know that the rule for f(x) = 1/x, and f(x) = y, and x = k, we know that f(x) = 1/x is the same as saying for a constant, k, f(k) = 1/k or as saying y = 1/k.