Implicit Derivatives - Trigonometry

In summary, the conversation is about a math problem involving finding the first and second derivative, critical values, points of inflection, intercepts, domain, range, asymptotes, and symmetry for the equation xsin(y) + x^3 = tan^(-1)(y). The process involves using Leibniz notation for implicit differentiation and applying the chain and product rules. The second derivative can also be found using the same process. The domain of the function is all real numbers, while the y-intercept is found by setting all x-values to zero and solving for y, and the x-intercept is found by setting all y-values to zero and solving for x. There are no asymptotes in the function.
  • #1
einsteinmic
2
0
Hello hello everyone...
I have a problem that I've been trying to figure out for a couple of days now. Its a question off of an assignment, and I have been busting my balls(excuse my french) with no results. HELP would be GREATLY appreciated...lthis is the question at hand:

xsin(y) + x^3 = tan^(-1)(y)
the second part is an inverse


I have to find the first and second derivative, critical values, points of inflection, intercept, in order to sketch it afterwards.

I shall cross my fingers with the hope that I will find genius out there :D !
 
Physics news on Phys.org
  • #2
use leibniz notation d/dx everything.
if you d(x^3)/dx , u get 3x^2.
The implicit part is differentiating the ys.If you assumme sin (y) = u --> du/dx = d(sin(y))/dx , is that clear?

now du/dx = du/dy . dy/dx becuase u is composed of the variable y. so if you do this you'll have du/dx = d(sin(y))/y . dy/dx. Leave dy/dx and d(sin(y))/y is cos(y).
so d(sin(y))/dx = cos(y) . dy/dx.

The rest is just applications of this, with product rule and everything. arctan (the inverse tan) has a derivative of (1/(1+x^2)). If you understand the chain rule it should be easy. Once you take the first derivative of everything, you can isolate the equation for dy/dx to be on one side and everything on the other to find critical values.

If you have any specific questions feel free to ask, good luck.
 
  • #3
Hey...
thanks for the reply it makes a lot of sense
The last question I have now is... how would I go about finding the x-y intercepts, domain, range, asymptotes and symmetry of the function??
By getting the derivative, I can find the critical values, local maximums/minimums, inflection points, and concavity intervals... but I also need to sketch the curve and in order to do so I would need the preceding info as well. I don't know how to isolate the y on one side, and the x on the other in order to do so :(

HEEEEELPP!

ps. Would I find the second derivative applying the same procedure (implicit differentiation)??

Your help is greatly GREATLY appreciated! :D
 
  • #4
The second derivative can be found the same process, just d(whatever is on each other side) by dx.

Let's say you have x^3 + y^3 = xy

First derivative is : 3x^2 + 3y^2 . dy/dx = y + x . dy/dx
Now if everything on left side is g(x) and the right side is n(x). Then just do dg(x) / dx and do dn(x) / dx.

So it would be : d[3x^2 + 3y^2 . dy/dx] / dx = d[y + x . dy/dx] / dx

6x + 3y^2 . (d^2y / dx^2) + 6y . (dy/dx) ^ 2 = 2 . dy/dx + x . (d^2y / dx^2)

That is the second derivative. Notice (dy/dx) ^ 2 is not same as (d^2y / dx^2).

One means the square of first derivative, second is notation for second derivative.

Domain is a concept, has no equations. Ok the first term has a sin, and all real numbers are ok in sin functions so you're good. The second term , x^3 not a problem, all real numbers. The right hand side has a tan^-1 which means it's domain would be from one asymptote of tan(x) to another. So y has to be from -pi/2 to pi/2 and that is the domain of y values in the function.

X-int means when y equals to 0, what can x be. Just replace all ys with zero.

Y-int is opposite, replace all xs with zero, solve for y.

I don't think there is any asymptotes in the function tho, there is no denominators.
 
Last edited:

1. What is the definition of an implicit derivative in trigonometry?

An implicit derivative in trigonometry is a type of derivative that is used to find the rate of change of a function that is not explicitly defined in terms of its independent variable. This means that the function cannot be easily rewritten in a form where the independent variable is isolated on one side of the equation.

2. How is an implicit derivative different from an explicit derivative in trigonometry?

An explicit derivative in trigonometry involves finding the derivative of a function where the independent variable is explicitly written on one side of the equation. An implicit derivative, on the other hand, involves finding the derivative of a function where the independent variable is not explicitly written and may be present on both sides of the equation.

3. What are the steps for finding an implicit derivative in trigonometry?

The steps for finding an implicit derivative in trigonometry are as follows:

  • Differentiate both sides of the equation with respect to the independent variable
  • Use the chain rule and product rule as needed
  • Isolate the derivative term on one side of the equation
  • Solve for the derivative

4. What are some common examples of implicit derivatives in trigonometry?

Some common examples of implicit derivatives in trigonometry include finding the derivative of trigonometric functions such as sine, cosine, and tangent, as well as inverse trigonometric functions such as arcsine, arccosine, and arctangent. These functions often involve both the independent variable and the trigonometric function on both sides of the equation.

5. How are implicit derivatives in trigonometry used in real-world applications?

Implicit derivatives in trigonometry are used in a variety of real-world applications, such as calculating the rate of change of a function in physics problems involving motion, finding the slope of a curve in engineering applications, and determining optimal solutions in economics and finance problems. They are also commonly used in the fields of calculus, physics, and engineering to solve complex mathematical equations and model real-world phenomena.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
928
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
896
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top