Question

Multiple Choice Which of the following is equal to the slope of the tangent to \(y^{2}-x^{2}=1\) at \((1, \sqrt{2}) ?\) ? \((\mathbf{A})-\frac{1}{\sqrt{2}} \quad(\mathbf{B})-\sqrt{2}\) \((\mathbf{C}) \frac{1}{\sqrt{2}} \quad(\mathbf{D}) \sqrt{2} \quad(\mathbf{E}) 0\)

Short answer

The correct multiple choice answer is \((\mathbf{C}) \frac{1}{\sqrt{2}}\)

Step-by-step solution

Differentiation

Differentiate the function \(y^{2}-x^{2}=1\) with respect to \(x\) using the chain rule. The chain rule is \(d(uv) = vdu + udv\). \nFirst, rewrite \(y^{2}-x^{2} = 1\) as \(y^{2} = x^{2} + 1\) to make differentiation easier. Then differentiate each side with respect to \(x\) to get \(2y dy/dx = 2x\). Solving for dy/dx gives \(dy/dx = x/y\).

Substitution

Now substitute \(x = 1\) and \(y = \sqrt{2}\) into the derivative to find the slope of the tangent line. This gives \(dy/dx = 1 / \sqrt{2}\).

Multiple Choice Selection

Finally, verify the solution by finding it among the multiple choices given.