Product and Quotient Rule — Question 1

Question 1

Let $$f(x) = x^2 \cdot \ln(x)$$

• Differentiate $f(x)$ using the product rule.

• Find the equation of the tangent line to the graph of $f(x)$ at $x = 1$.

Question 1 - Solution

We are given: $$f(x) = x^2 \cdot \ln(x)$$

Let: $$u = x^2, \quad v = \ln(x) \Rightarrow u' = 2x, \quad v' = \frac{1}{x}$$

Apply the product rule: $$f'(x) = u'v + uv' = 2x \ln(x) + x^2 \cdot \frac{1}{x} = 2x \ln(x) + x$$

Final derivative: $$\boxed{f'(x) = 2x \ln(x) + x}$$

Now evaluate the tangent line at $x = 1$:

$$f(1) = 1^2 \cdot \ln(1) = 0$$ $$f'(1) = 2(1)\ln(1) + 1 = 0 + 1 = 1$$

Point on the curve: $(1, 0)$ Slope of tangent: $1$

Use point-slope form: $$y - 0 = 1(x - 1) \Rightarrow y = x - 1$$

Equation of the tangent line: $$\boxed{y = x - 1}$$