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Ind \( \frac{d y}{d x} \) if \( x=a \sin ^{2} t \) and \( y=\operatorname{aros}^{2} t, a \) is \( n \) const

Ask by Chang Gordon. in Nigeria
Mar 11,2025

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Answer

The derivative \( \frac{dy}{dx} \) is \( \frac{\arcsin t}{a \sin t \cos t \sqrt{1 - \sin^2 t}} \).

Solution

It seems that there was an issue with the calculation. Let's try to find the first order derivative of \( y \) with respect to \( x \) again. Given that \( x = a \sin^2 t \) and \( y = \arcsin^2 t \), we can find the derivative of \( y \) with respect to \( x \) using the chain rule. First, let's find the derivative of \( y \) with respect to \( t \): \[ \frac{dy}{dt} = 2 \arcsin t \cdot \frac{1}{\sqrt{1 - \sin^2 t}} \] Next, let's find the derivative of \( x \) with respect to \( t \): \[ \frac{dx}{dt} = 2a \sin t \cos t \] Now, we can use the chain rule to find the derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \] Substitute the expressions for \( \frac{dy}{dt} \) and \( \frac{dt}{dx} \) into the chain rule formula: \[ \frac{dy}{dx} = 2 \arcsin t \cdot \frac{1}{\sqrt{1 - \sin^2 t}} \cdot \frac{1}{2a \sin t \cos t} \] Simplify the expression: \[ \frac{dy}{dx} = \frac{\arcsin t}{a \sin t \cos t \sqrt{1 - \sin^2 t}} \] Therefore, the derivative of \( y \) with respect to \( x \) is \( \frac{\arcsin t}{a \sin t \cos t \sqrt{1 - \sin^2 t}} \).

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The Deep Dive

To find \( \frac{d y}{d x} \), we will need to use the chain rule. First, let's denote \( x = a \sin^2 t \) and \( y = \operatorname{aros}^2 t \). We start by differentiating both \( x \) and \( y \) with respect to \( t \): 1. For \( x \): \[ \frac{dx}{dt} = a \cdot 2 \sin t \cos t = a \sin(2t) \] 2. For \( y \): \[ \frac{dy}{dt} = 2 \operatorname{aros}(t) \cdot \frac{d (\operatorname{aros}(t))}{dt} \] Since \( \frac{d(\operatorname{aros}(t))}{dt} \) depends on the function you are considering, you may refer to the derivative of the inverse function you have. Now, to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \operatorname{aros}(t) \cdot \frac{d(\operatorname{aros}(t))}{dt}}{a \sin(2t)} \] So the complete expression depends on how you differentiate \( \operatorname{aros}(t) \). If we assume it's a hypothetical function with a known derivative, plug that into the formula above: \[ \frac{dy}{dx} = \frac{2 \operatorname{aros}(t) \cdot \text{[known derivative]}}{a \sin(2t)} \] Make sure to calculate and substitute the appropriate values based on the specifics of \( \operatorname{aros}(t) \) if they are defined in your problem context.

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