Question

differentiate ( y = ln(10 - x - 5x^2) ).( y = )

Explanation:

Step1: Recall the chain rule for differentiation.

The chain rule states that if \( y = \ln(u) \) where \( u \) is a function of \( x \), then \( y'=\frac{1}{u}\cdot u' \). Here, \( u = 10 - x - 5x^{2} \).

Step2: Differentiate \( u \) with respect to \( x \).

Differentiate \( u = 10 - x - 5x^{2} \) term - by - term. The derivative of a constant (10) is 0, the derivative of \( -x \) with respect to \( x \) is \( - 1\), and the derivative of \( -5x^{2}\) with respect to \( x \) is \( - 10x \) (using the power rule \( \frac{d}{dx}(ax^{n})=nax^{n - 1} \)). So, \( u'=\frac{d}{dx}(10 - x - 5x^{2})=-1-10x \).

Step3: Apply the chain rule.

Since \( y=\ln(u) \) and \( u = 10 - x - 5x^{2} \), then \( y'=\frac{1}{u}\cdot u'=\frac{-1 - 10x}{10 - x - 5x^{2}} \). We can factor out a negative sign from the numerator: \( y'=\frac{-(1 + 10x)}{10 - x - 5x^{2}}=\frac{10x + 1}{5x^{2}+x - 10} \) (by multiplying the numerator and denominator by - 1 to make the denominator in a more standard form).

Answer:

\( y'=\frac{-1 - 10x}{10 - x - 5x^{2}} \) (or \( \frac{10x + 1}{5x^{2}+x - 10} \))