Question

In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.35. y = ln((x²$$+1)^5$$/√(1-x))

Accepted Answer

Rewrite the function to simplify differentiation by expressing the logarithm of a quotient and powers as sums and differences: use the property $$\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$$ and $$\ln(a^n) = n \ln(a). So $$rewrite $$y = \ln\left(\frac{(x^2$+1)^5$$}{\sqrt{1-x}}\right)$$ as y = 5$$ \$$ln(x^2 + 1$$) - \frac{1}{2} \ln(1 - x)$$. Differentiate each term separately with respect to x$ using the chain rule. For the first term, 5$$ \$$ln(x^2 + 1$$)$$, the derivative is 5$$ \cdot \frac{1}{$$x^2 + 1$$} \cdot \frac{d}{dx}$$(x^2 + 1$$)$$. Calculate the derivative inside the chain rule for the first term: $$\frac{d}{dx}$$(x^2 + 1) = 2x$. For the second term, $$-\frac{1}{2} \ln(1 - x)$$, differentiate using the chain rule: $$-\frac{1}{2} \cdot \frac{1}{1 - x} \cdot \frac{d}{dx}(1 - x)$$. Calculate the derivative inside the chain rule for the second term: $$\frac{d}{dx}(1 - x) = -1$$. Then combine all parts to write the derivative $$\frac{dy}{dx}$ as the sum of these differentiated terms.

In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
35. y = ln((x²+1)^5/√(1-x))

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Key Concepts

Logarithmic Differentiation

Chain Rule

Properties of Logarithms

In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.35. y = ln((x²+1)^5/√(1-x))