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Exercise 10.2 · Q17

Q.Find the derivative of the following function with respect to the corresponding independent variable: y=(x2+5)log⁡(1+x) e−3xy = (x^2+5)\log(1+x)\,e^{-3x}

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Step 1. Let u=x2+5u = x^2+5, v=log⁡(1+x)v = \log(1+x), w=e−3xw = e^{-3x}. Then u′=2xu' = 2x, v′=11+xv' = \dfrac{1}{1+x} (chain rule), w′=−3e−3xw' = -3e^{-3x} (chain rule).

Step 2. Apply the extended product rule y′=u′vw+uv′w+uvw′y' = u'vw + uv'w + uvw':

y′=2xlog⁡(1+x)e−3x+(x2+5)⋅11+x⋅e−3x+(x2+5)log⁡(1+x)(−3e−3x)y' = 2x\log(1+x)e^{-3x} + (x^2+5)\cdot\dfrac{1}{1+x}\cdot e^{-3x} + (x^2+5)\log(1+x)(-3e^{-3x}).

Step 3. Factor out e−3xe^{-3x}: y′=e−3x[2xlog⁡(1+x)+x2+51+x−3(x2+5)log⁡(1+x)]y' = e^{-3x}\left[2x\log(1+x) + \dfrac{x^2+5}{1+x} - 3(x^2+5)\log(1+x)\right]. …

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