Find The Second Derivative Value
Alright math whizzes, let's dive into a super interesting calculus problem that'll test your skills with derivatives. We've got a twice-differentiable function, let's call it $y=f(x)$, and we know a couple of key things about it. First off, when $x=1$, the function's value is $f(1)=2$. Pretty straightforward, right? Now, here's where it gets a bit spicy: the first derivative, $ rac{d y}{d x}$, is given by the expression $y^3+3$. Our mission, should we choose to accept it, is to find the value of the second derivative, $ rac{d^2 y}{d x^2}$, specifically at $x=1$. This is like peeling back layers of a mathematical onion, guys!
To get to the bottom of this, we need to employ the power of implicit differentiation, and then differentiate again. Since we're given $ racd y}{d x} = y^3+3$, we need to differentiate both sides with respect to $x$ to find $ rac{d^2 y}{d x^2}$. Remember, $y$ is a function of $x$, so we'll need to use the chain rule. Differentiating $y^3$ with respect to $x$ gives us $3y^2 rac{d y}{d x}$. And the derivative of a constant, $3$, is just $0$. So, $ rac{d^2 y}{d x^2} = 3y^2 rac{d y}{d x}$. See? We've found an expression for the second derivative in terms of $y$ and $ rac{d y}{d x}$. Now, we just need to plug in the values we know. We are given that at $x=1$, $y=2$. We also know that $ rac{d y}{d x} = y^3+3$. So, at $x=1$, we can calculate $ rac{d y}{d x}$d x}igg|_{x=1} = (f(1))^3 + 3 = 2^3 + 3 = 8 + 3 = 11$. Awesome! Now we have all the pieces to find $ rac{d^2 y}{d x^2}$ at $x=1$. Plugging in $y=2$ and $ rac{d y}{d x}=11$ into our expression for the second derivative, we get{d x^2}igg|{x=1} = 3(f(1))^2 rac{d y}{d x}igg|{x=1} = 3(2)^2 (11) = 3(4)(11) = 12 imes 11 = 132$. So, the value of the second derivative at $x=1$ is 132. Pretty neat, huh? It's all about breaking down the problem step-by-step and applying the right calculus rules. Keep practicing, and you'll be a derivative master in no time!
Let's recap the journey we just took, guys. We started with a function $y=f(x)$ that's twice-differentiable. The initial conditions were $f(1)=2$ and the first derivative $ racd y}{d x}=y^3+3$. Our goal was to find $ rac{d^2 y}{d x^2}$ at $x=1$. The first crucial step was to differentiate the given expression for $ rac{d y}{d x}$ with respect to $x$. Using the chain rule because $y$ is a function of $x$, we found that $ rac{d^2 y}{d x^2} = rac{d}{dx}(y^3+3)$. This breaks down into $ rac{d}{dx}(y^3) + rac{d}{dx}(3)$. The derivative of $3$ is $0$. For $y^3$, we apply the chain ruled x}$. Thus, we arrived at the expressiond x^2} = 3y^2 rac{d y}{d x}$. Now, we needed to evaluate this at the specific point $x=1$. We were given $f(1)=2$, which means $y=2$ when $x=1$. We also needed the value of the first derivative at $x=1$. We could find this using the given equation $ rac{d y}{d x}=y^3+3$. Substituting $y=2$ into this, we get $ rac{d y}{d x}igg|_{x=1} = 2^3 + 3 = 8 + 3 = 11$. With both $y$ and $ rac{d y}{d x}$ at $x=1$ in hand, we could finally calculate the second derivative. Plugging $y=2$ and $ rac{d y}{d x}=11$ into our formula for $ rac{d^2 y}{d x^2}${d x^2}igg|_{x=1} = 3(2)^2 (11) = 3(4)(11) = 12 imes 11 = 132$. And there you have it – the value of the second derivative at $x=1$ is 132. This problem is a classic example of how implicit differentiation and the chain rule work hand-in-hand to solve complex calculus problems. Keep practicing these techniques, and you'll master them in no time! The options provided were (A) 12, (B) 66, (C) 132, and (D) 165. Our calculated value matches option (C). Nicely done, everyone!
So, you've seen how we tackle this kind of problem, but let's really break down why each step is crucial. The initial information, $f(1)=2$ and $ racd y}{d x}=y^3+3$, are our starting blocks. The fact that $y=f(x)$ is twice-differentiable is a green light; it tells us that $ rac{d^2 y}{d x^2}$ actually exists and we can find it. The first derivative, $ rac{d y}{d x}=y^3+3$, is the engine of our problem. It relates the rate of change of $y$ with respect to $x$ to the current value of $y$. When we want to find the second derivative, $ rac{d^2 y}{d x^2}$, we are essentially asking about the rate of change of the first derivative. This is where implicit differentiation becomes our best friend. We take the derivative of $ rac{d y}{d x}=y^3+3$ with respect to $x$. On the left side, $ rac{d}{dx}igg( rac{d y}{d x}igg)$ is simply $ rac{d^2 y}{d x^2}$. On the right side, $ rac{d}{dx}(y^3+3)$, we have to be careful. The derivative of $+3$ is $0$, but the derivative of $y^3$ requires the chain rule. Think of $y^3$ as $(g(x))^3$, where $g(x)=y$. The chain rule states that the derivative is $3(g(x))^2 imes g'(x)$, which translates to $3y^2 rac{d y}{d x}$. So, $ rac{d^2 y}{d x^2} = 3y^2 rac{d y}{d x}$. Now, we need to evaluate this at $x=1$. This means we need the values of $y$ and $ rac{d y}{d x}$ at $x=1$. We are given $f(1)=2$, so $y=2$ at $x=1$. To find $ rac{d y}{d x}$ at $x=1$, we substitute $y=2$ into the given first derivative equationd x}igg|_{x=1} = (f(1))^3 + 3 = 2^3 + 3 = 8 + 3 = 11$. Now we have everything! Substitute $y=2$ and $ rac{d y}{d x}=11$ into our expression for $ rac{d^2 y}{d x^2}${d x^2}igg|_{x=1} = 3(2)^2 (11) = 3(4)(11) = 12 imes 11 = 132$. This result, 132, is our answer. The key takeaways here are understanding what the second derivative represents (the rate of change of the rate of change) and mastering the chain rule during implicit differentiation. It’s like building a higher-order structure from the fundamental building blocks of the function and its first derivative. Keep at it, guys, and these concepts will become second nature!