Question

1. let ( y = f(x) ) be a twice - differentiable function such that ( f(1)=2 ) and ( \frac{d y}{d x}=y^{2}+3 ). what is the value of ( \frac{d^{2} y}{d x^{2}} ) at ( x = 1 )?

Explanation:

Step1: Recall the formula for the second derivative using implicit differentiation

To find \(\frac{d^2y}{dx^2}\), we start with the given first derivative \(\frac{dy}{dx}=y^3 + 3\). We use the chain rule (implicit differentiation) to differentiate both sides with respect to \(x\). The derivative of \(\frac{dy}{dx}\) with respect to \(x\) is \(\frac{d^2y}{dx^2}\), and the derivative of \(y^3+3\) with respect to \(x\) is \(3y^2\frac{dy}{dx}+0\) (by the chain rule: derivative of \(y^3\) with respect to \(y\) is \(3y^2\), then multiply by \(\frac{dy}{dx}\) to get the derivative with respect to \(x\), and the derivative of 3 with respect to \(x\) is 0). So we have:
\(\frac{d^2y}{dx^2}=3y^2\frac{dy}{dx}\)

Step2: Substitute the value of \(y\) at \(x = 1\) and the value of \(\frac{dy}{dx}\) at \(x = 1\)

We know that \(y=f(x)\) and \(f(1) = 2\), so when \(x = 1\), \(y=2\). Also, when \(x = 1\), \(\frac{dy}{dx}=y^3+3\). Substituting \(y = 2\) into the first - derivative formula: \(\frac{dy}{dx}\big|_{x = 1}=2^3+3=8 + 3=11\)

Now substitute \(y = 2\) and \(\frac{dy}{dx}=11\) (at \(x = 1\)) into the formula for \(\frac{d^2y}{dx^2}\):
\(\frac{d^2y}{dx^2}\big|_{x = 1}=3\times(2)^2\times11\)

First, calculate \(2^2 = 4\), then \(3\times4=12\), and finally \(12\times11 = 132\)

Answer:

\(132\)