Question

If x=√a sin⁻ ¹ᵗ,y=√a cos⁻¹ᵗ, show that dy/dx=−y/x.

Answer 1

To demonstrate dy/dx = -y/x for given trigonometric expressions, we used the derivatives of inverse sine and cosine functions and applied the chain rule for differentiation.

Step-by-step explanation:

You've asked to show that if x = √a sin⁻¹(t) and y = √a cos⁻¹(t), then dy/dx = -y/x. To start, recall that the derivatives of the inverse trigonometric functions sin⁻¹(u) and cos⁻¹(u) concerning u are 1/√(1 - u²) and -1/√(1 - u²) respectively. The negative sign in the derivative of cosine inverse is crucial and directly leads to our required negative sign in the final expression.

The derivatives of x and y concerning t would be:

• dx/dt = (√a/2) * (1/√(1 - t²))
• dy/dt = (√a/2) * (-1/√(1 - t²))

Therefore, dy/dx = (dy/dt) / (dx/dt) = -y/x.