Differentiation Of Inverse Function

In summary, the problem is asking to find the derivative of y = 1/a * tan^-1(x/a). The solution is to use the chain rule and the derivative of inverse tangent. The final answer is 1/(a^2 + x^2), but the incorrect answer was given as 1/(x^2 + a). The error was using the incorrect formula for the derivative of inverse tangent.
  • #1
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Homework Statement


[itex]\frac{\mathrm{d}\left(\frac{1}{a}\tan ^{-1} \left(\frac{x}{a}\right)\right)}{\mathrm{d}x}[/itex]2. The attempt at a solution
Let [itex]y = \frac{1}{a}\tan ^{-1} \left(\frac{x}{a}\right)[/itex]
[itex]\therefore x = a \tan \left(ay\right)[/itex]
Differentiate with respect to [itex]x[/itex] [itex]\rightarrow 1 = a \sec ^2 \left(ay\right) \frac{\mathm{d}y}{\mathrm{d}x}[/itex]
[itex]\therefore \frac{\mathm{d}y}{\mathrm{d}x} = \frac{1}{a \sec ^2 \left(ay\right)} = \frac{1}{a \tan ^2 \left(ay\right) + a}[/itex]
[itex]x = a \tan \left(ay\right) \therefore \frac{1}{a \tan ^2 \left(ay\right) + a} = \frac{1}{\left(a \tan \left(ay\right)\right)^2 + a} = \frac{1}{x^2 + a}[/itex]3. The problem I encountered
However, thee answer is incorrect. The correct answer is:
[itex]= \frac{1}{a^2+x^2}[/itex]
Where have I gone wrong?
 
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  • #2
x=a tan(ay)
1= a sec^2(ay) d(ay)/dx
 

1. What is the inverse function?

The inverse function is a mathematical operation that reverses the effect of another function. It is denoted as f^-1 and can be thought of as "undoing" the original function.

2. How is the inverse function related to differentiation?

The inverse function and differentiation are closely related because the derivative of an inverse function is the reciprocal of the derivative of the original function. This means that if the derivative of a function f(x) is g(x), then the derivative of its inverse function f^-1(x) is 1/g(x).

3. What is the process for differentiating an inverse function?

The process for differentiating an inverse function involves using the inverse function theorem, which states that the derivative of an inverse function is equal to 1 divided by the derivative of the original function evaluated at the inverse function's input.

4. What is the importance of differentiating inverse functions?

Differentiating inverse functions is important because it allows us to find the slope or rate of change of the original function at a particular point. This can be useful in various real-world applications, such as finding the velocity of an object or the growth rate of a population.

5. Are there any special cases when differentiating inverse functions?

Yes, there are two special cases when differentiating inverse functions. The first case is when the original function has a vertical tangent line at a point, in which case the inverse function will have a horizontal tangent line at that same point. The second case is when the original function is not one-to-one, meaning it has more than one output for a given input. In this case, the inverse function does not exist or has multiple branches.

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