Differentiation Formulas — Question 8

Question 8 - Solution

(a) Simplify the Function:

We start with: $$f(x) = \ln\left( \sqrt{x^2 + 3x + 2} \right)$$

Since $$\sqrt{x^2+3x+2} = (x^2+3x+2)^{1/2},$$ we can write: $$f(x) = \ln\left((x^2+3x+2)^{1/2}\right)$$

Using the logarithm rule $$\ln(A^r) = r\ln(A),$$ we get: $$f(x) = \frac{1}{2}\ln(x^2+3x+2)$$

So the simplified function is: $$\boxed{f(x) = \frac{1}{2}\ln(x^2+3x+2)}$$

(b) Find $f'(x)$:

Now differentiate: $$f(x) = \frac{1}{2}\ln(x^2+3x+2)$$

Using $$\frac{d}{dx}\ln(u) = \frac{u'}{u},$$ where $$u = x^2+3x+2$$ and $$u' = 2x+3,$$ we get: $$f'(x) = \frac{1}{2}\cdot \frac{2x+3}{x^2+3x+2}$$

Therefore, $$\boxed{f'(x) = \frac{2x+3}{2(x^2+3x+2)}}$$