Problem that uses implicit differentiation

[gigi9]

Can u please give me an example of a problem that use implicit differentiation to find the derivative of an inverse function and of relations. Plz also explain to me what relation means in this case. Thanks
also please give me an example that use the definite integral to compute accumulated changes.
Thanks alot

 

[Hurkyl]

Suppose you wanted to find the derivative of ln x, which is the inverse of e^x.

e^{ln x} = x

differentiate both sides WRT x

e^{ln x} (ln x)' = 1
x (ln x)' = 1
(ln x)' = 1 / x

so the derivative of ln x is 1/x


A relation is (to slightly modify the definition to make it easier to understand) simply a mapping from pairs of numbers to "true" or "false".

For instance:
x < y
is a relation.

Also, for any function f:
f(x) = y
is a relation. In fact, all functions are relations. They just have the special property that:
f(x) = y and f(x) = z
can both be true only if y = z.


An example of a relation where one might use implicit differentiation is to find the slope of a point on the unit circle.

x² + y² = 1

is a relation. Notice that y cannot be written as a function of x because for some values of x, there are two solutions for y... but we can still differentiate WRT x to get:

2 x + 2 y (dy/dx) = 0
2 y (dy/dx) = - 2x
(dy/dx) = -x / y

So, the slope of the tangent line to any point (x, y) of the unit circle is (dy/dx) = -x / y

Thanks
also please give me an example that use the definite integral to compute accumulated changes.

Suppose I throw a ball into the air straight up with a speed of 19.6 meters per second. How high is it after 2 seconds?

Well, acceleration is simply the rate change of velocity, so we can find the velocity at any time t with a definite integral:

v(t) - v(0) = &int_s=0..t a(s) ds
v(t) - 19.6 = &int_s=0..t -9.8 ds
v(t) = 19.6 + (-9.8) * (t - 0)
v(t) = 19.6 - 9.8 t

Now, velocity is simply the rate chance of position, so we can find the height at time 2 with a definite integral:
x(2) - x(0) = &int_s=0..2 v(s) ds
x(2) - 0 = &int_s=0..2 19.6 - 9.8 s ds
x(2) = [19.6 s - 4.9 s²]_s=0..2
x(2) = (19.6 * 2 - 4.9 * 2²) - (19.6 * 0 - 4.9 * 0²)
x(2) = 19.6

So after 2 seconds, the ball is 19.6 meters high.

 

[tacman]

Sure, I'd be happy to provide an example for each of these cases.

1. Implicit Differentiation:
Problem: Find the derivative of the inverse function of f(x) = x^2 + 2x + 3.

Solution:
To find the derivative of the inverse function, we first need to find the inverse function itself. So, let's solve for x in terms of y:
y = x^2 + 2x + 3
x^2 + 2x + (3 - y) = 0
Using the quadratic formula, we get:
x = (-2 ± √(4 - 4(3-y))) / 2
= (-2 ± √(12 - 4y)) / 2
= (-1 ± √(3 - y)) / 1

Therefore, the inverse function is f^-1(y) = (-1 ± √(3 - y)) / 1

Now, to find the derivative of the inverse function, we use implicit differentiation. Let's differentiate both sides of the equation with respect to y:
d/dy(f^-1(y)) = d/dy((-1 ± √(3 - y)) / 1)
= d/dy(-1) ± d/dy(√(3 - y)) / d/dy(1)
= 0 ± (-1/2√(3 - y)) * (-1)
= 1/2√(3 - y)

So, the derivative of the inverse function is f^-1'(y) = 1/2√(3 - y).

Relation in this case refers to the relationship between two variables, x and y, in an equation. In the above example, the relation is y = x^2 + 2x + 3, where y is a function of x.

2. Definite Integral:
Problem: Find the accumulated change of the function f(x) = 2x^2 + 3x from x = 1 to x = 5.

Solution:
To find the accumulated change, we need to use the definite integral. The definite integral represents the area under the curve of a function between two given points. So, we can write the definite integral as:
∫(1 to 5) f(x) dx

Plugging in the function f(x) = 2x^2 + 3x, we