Questions: y=x^2 ln x, (1,0)

Transcript text: $y=x^{2} \ln x, \quad(1,0)$

Solution

Solution Steps

To find the derivative of the function $ y = x^2 \ln x $ at the point $ (1, 0) $, we will use the product rule and the chain rule. The product rule states that the derivative of a product of two functions $ u(x) $ and $ v(x) $ is given by $ u'(x)v(x) + u(x)v'(x) $. Here, we can let $ u(x) = x^2 $ and $ v(x) = \ln x $. We will find the derivatives $ u'(x) $ and $ v'(x) $, apply the product rule, and then evaluate the derivative at $ x = 1 $.

Step 1: Define the Function

We start with the function given by $ y = x^2 \ln x $.

Step 2: Differentiate the Function

To find the derivative $ \frac{dy}{dx} $, we apply the product rule. Let $ u = x^2 $ and $ v = \ln x $. Then, we have: \[ \frac{dy}{dx} = u'v + uv' = (2x)(\ln x) + (x^2)\left(\frac{1}{x}\right) = 2x \ln x + x \]

Step 3: Evaluate the Derivative at $ x = 1 $

Now, we evaluate the derivative at the point $ x = 1 $: \[ \frac{dy}{dx} \bigg|_{x=1} = 2(1) \ln(1) + 1 = 2(1)(0) + 1 = 1 \]