Question

In Exercises \(1-8,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\sin ^{-1} \frac{3}{t^{2}}$$

Short answer

The derivative of \(y = \sin^{-1}(\frac{3}{t^{2}})\) with respect to \(t\) is \(y' = -\frac{6}{t^{3} \sqrt{1 - (\frac{9}{t^{4}})}}\).

Step-by-step solution

Differentiation of inverse sine

Recall the formula for differentiating the inverse sine function \(y = \sin^{-1}(x)\): \\ The derivative is given by \(\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^{2}}}\).

Use Chain Rule

The function is a composition of the inverse sine function \(\sin^{-1}(x)\) and the function \(\frac{3}{t^{2}}\). This means that the chain rule is needed to differentiate it: \\ The chain rule states that the derivative of the composite function \(f(g(x))\) is given by \(f'(g(x)) \cdot g'(x)\). In this case, \(f(x) = \sin^{-1}(x)\) and \(g(x) = \frac{3}{t^{2}}\). \\ Differentiating the outer function, \(f'(x) = \frac{1}{\sqrt{1 - x^{2}}}\), and substituting \(g(x)\) into it, gives \(f'(g(x)) = \frac{1}{\sqrt{1 - (\frac{3}{t^{2}})^{2}}}\).

Differentiate the Inner Function

The inner function \(g(x) = \frac{3}{t^{2}}\) is a simple power function, and so can be differentiated using the power rule. Its derivative is therefore \(g'(x) = -\frac{6}{t^{3}}\).

Substitute the Derivatives into the Chain Rule

Substituting the values of \(f'(g(x))\) and \(g'(x)\) into the chain rule formula, gives: \\ \(y' = f'(g(x)) \cdot g'(x) = \frac{1}{\sqrt{1 - (\frac{3}{t^{2}})^{2}}} \cdot -\frac{6}{t^{3}} = -\frac{6}{t^{3} \sqrt{1 - (\frac{9}{t^{4}})}}\).