If u = sin^-1(x) + tan^-1(y), find the value of x * (du/dx) + y * (du/dy) = ? Prove that (x * du/dx) - (y * du/dy) = ...

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Understand the Problem

The question appears to involve finding derivatives of a function using implicit differentiation and proving a relationship involving partial derivatives. It asks for the derivatives of a specific function related to inverse trigonometric functions.

Answer

1. \( x \frac{du}{dx} + y \frac{du}{dy} = \frac{x}{\sqrt{1 - x^2}} + \frac{y}{1 + y^2} \) 2. \( x \frac{du}{dx} - y \frac{du}{dy} = \frac{x}{\sqrt{1 - x^2}} - \frac{y}{1 + y^2} \)
Answer for screen readers

The values calculated are:

  1. ( x \frac{du}{dx} + y \frac{du}{dy} = \frac{x}{\sqrt{1 - x^2}} + \frac{y}{1 + y^2} )

  2. ( x \frac{du}{dx} - y \frac{du}{dy} = \frac{x}{\sqrt{1 - x^2}} - \frac{y}{1 + y^2} )

Steps to Solve

  1. Find the Derivatives of ( u )

Given ( u = \sin^{-1}(x) + \tan^{-1}(y) ), we need to find ( \frac{du}{dx} ) and ( \frac{du}{dy} ).

Using the derivatives:

  • The derivative of ( \sin^{-1}(x) ) is ( \frac{1}{\sqrt{1 - x^2}} ).
  • The derivative of ( \tan^{-1}(y) ) is ( \frac{1}{1 + y^2} ).

So we have:

$$ \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} + 0 = \frac{1}{\sqrt{1 - x^2}} $$

$$ \frac{du}{dy} = 0 + \frac{1}{1 + y^2} = \frac{1}{1 + y^2} $$

  1. Calculate ( x \frac{du}{dx} + y \frac{du}{dy} )

Now substitute the derivatives into the expression ( x \frac{du}{dx} + y \frac{du}{dy} ):

[ x \frac{du}{dx} + y \frac{du}{dy} = x \left(\frac{1}{\sqrt{1 - x^2}}\right) + y \left(\frac{1}{1 + y^2}\right) = \frac{x}{\sqrt{1 - x^2}} + \frac{y}{1 + y^2} ]

  1. Proving ( (x \frac{du}{dx}) - (y \frac{du}{dy}) )

Now we find ( x \frac{du}{dx} - y \frac{du}{dy} ):

[ x \frac{du}{dx} - y \frac{du}{dy} = \frac{x}{\sqrt{1 - x^2}} - \frac{y}{1 + y^2} ]

This is the form we need to prove to find its value.

The values calculated are:

  1. ( x \frac{du}{dx} + y \frac{du}{dy} = \frac{x}{\sqrt{1 - x^2}} + \frac{y}{1 + y^2} )

  2. ( x \frac{du}{dx} - y \frac{du}{dy} = \frac{x}{\sqrt{1 - x^2}} - \frac{y}{1 + y^2} )

More Information

The derivatives of the inverse trigonometric functions play a crucial role in differentiating composite functions. This problem combines knowledge of implicit differentiation with applications of derivatives in real-world situations.

Tips

  • Forgetting to include constant derivatives when taking derivatives for mixed variables.
  • Incorrectly applying the chain rule, especially for inverse trigonometric functions.

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