Derivative of a logarithm with a polynomial argument

Let $f(x) = \ln(5x^2 - 3x + 1)$ . Find $f'(x)$ .

Hint

Use the chain rule with $\ln(u)$ as the outer function and $u = 5x^2 - 3x + 1$ as the inner function: $\frac{d}{dx}[\ln u] = \frac{u'}{u}$ .

Step-by-step solution

Recognize that $f(x) = \ln(u)$ where $u = 5x^2 - 3x + 1$ . By the chain rule, $\frac{d}{dx}[\ln(u)] = \frac{1}{u}\cdot \frac{du}{dx}$ , so we need $u'$ first.

Compute $u' = \frac{d}{dx}[5x^2 - 3x + 1] = 10x - 3$ . Substituting back gives $f'(x) = \dfrac{10x - 3}{5x^2 - 3x + 1}$ .

Derivative of a logarithm with a polynomial argument

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