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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.2.35

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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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.

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