Implicit Derivative of y=31*arctan(x)

In summary, the correct approach to finding the derivative of the given function is to use the constant multiple rule instead of the product rule. Additionally, the derivative of arctan(x) is 1 / (1+x^2) and when using the chain rule, it is important to substitute the appropriate value for u in the derivative formula. Finally, if the title of the problem is "Implicit Derivative", then the correct approach is to use implicit differentiation, finding the derivative of both sides of the equation with respect to the variable.
  • #1
winslow
3
0

Homework Statement


Find the derivative of the following function. Simplify where possible.
y=31*arctan(sqrt(x))


Homework Equations


I know that the derivative of arctan(x) = 1 / (1+x2)
I also know we will be using chain rule and product rule.

The Attempt at a Solution



y' = (31)'[arctan(sqrt(x))] + (31)[arctan(sqrt(x))]'
y' = (0)[arctan(sqrt(x))] + (31)*[1/(1+x2)] * (2sqrt(x))

y' = 31 / (2sqrt(x))(1+x2)


However the correct answer is
y' = 31 / (2sqrt(x))(1+x) <-- no x2

I'm not sure why the x2 ends up being just x. I checked if it was simplifying problem but that wasn't it (at least not from what I see).


Thanks for the help in advance.
 
Physics news on Phys.org
  • #2
winslow said:

Homework Statement


Find the derivative of the following function. Simplify where possible.
y=31*arctan(sqrt(x))


Homework Equations


I know that the derivative of arctan(x) = 1 / (1+x2)
I also know we will be using chain rule and product rule.

The Attempt at a Solution



y' = (31)'[arctan(sqrt(x))] + (31)[arctan(sqrt(x))]'
You should never use the product rule when one factor of the product is a constant. In such cases, you should always use the constant multiple rule; i.e., if y = k*f(x), y' = k*f'(x). It's not that using the product rule will give you the wrong answer; it's just that it adds too much complication, increasing the chances of making a mistake.
winslow said:
y' = (0)[arctan(sqrt(x))] + (31)*[1/(1+x2)] * (2sqrt(x))
In the line above you have two mistakes. d/dx(arctan(f(x)) = 1/[1 + (f(x))^2] * f'(x). In your problem, f(x) = sqrt(x), so (f(x))^2 = x. Also, d/dx(sqrt(x)) != 2sqrt(x).
winslow said:
y' = 31 / (2sqrt(x))(1+x2)


However the correct answer is
y' = 31 / (2sqrt(x))(1+x) <-- no x2

I'm not sure why the x2 ends up being just x. I checked if it was simplifying problem but that wasn't it (at least not from what I see).


Thanks for the help in advance.
 
  • #3
when you use the chain rule, you're supposed to think of it in the sense [tex]\frac{dg(u)}{dx}=g'(u)*du/dx[/tex]
so for you're case

[tex]\frac{dg(u)}{dx}=\frac{d}{dx}(a*\arctan (u))=a*\frac{1}{1+u^{2}}*\frac{du}{dx}[/tex]

just plug in u=sqrt(x)
 
  • #4
In addition, the title of your thread is Implicit Derivative of ..., so you are probably not expected to know the derivative of arctan(x).

Assuming your title actually means something, you are probably expected to do something like this:

y = 31 arctan(sqrt(x)) ==> y/31 = arctan(sqrt(x)) ==> sqrt(x) = tan(y/31) ==> x = [tan(y/31)]^2. Now take the derivative (with respect to y) implicitly.
 
  • #5
Oh I see where I went wrong with the arctan(u) and u = sqrt(x)

Thank you Mark44 and tt2348 for helping me! I appreciate it!
 

Related to Implicit Derivative of y=31*arctan(x)

1. What is the formula for finding the implicit derivative of y=31*arctan(x)?

The formula for finding the implicit derivative of y=31*arctan(x) is d/dx(y) = 31/(1+x^2).

2. How do you solve for the implicit derivative of y=31*arctan(x)?

To solve for the implicit derivative of y=31*arctan(x), you need to use the chain rule and the derivative of arctan(x), which is 1/(1+x^2). The final result will be 31/(1+x^2).

3. What is the meaning of the implicit derivative in the context of y=31*arctan(x)?

The implicit derivative in the context of y=31*arctan(x) represents the rate of change of y with respect to x when both variables are related by the equation y=31*arctan(x).

4. How does the implicit derivative of y=31*arctan(x) relate to the slope of the tangent line?

The implicit derivative of y=31*arctan(x) is equivalent to the slope of the tangent line at any given point on the curve. This means that it represents the instantaneous rate of change of y with respect to x at that point.

5. Can the implicit derivative of y=31*arctan(x) be used to find the derivative of other functions?

Yes, the implicit derivative of y=31*arctan(x) can be used as a general formula to find the derivative of any function that can be written in the form y=f(x)*g(x), where f(x) and g(x) are known functions. It is a useful tool in finding derivatives of more complex functions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
358
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
173
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
538
Replies
9
Views
750
Replies
7
Views
547
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
575
Back
Top