My Math Forum combinding chain and power rule

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 October 5th, 2016, 08:11 AM #1 Newbie   Joined: Oct 2016 From: Norway Posts: 1 Thanks: 0 combinding chain and power rule Hi, I have some trouble with an expression. The expression i'm supposed to solve is find the derivative of $\displaystyle f(x)=(x^(sin(x)))*(x2+1)^(-3x)$ d/dx u(x)^(v(x)) = ?? found something on an online calculator suggesting that the formula for a chain powered by a chain is: d/dx u(x)^(v(x)) = u(x)^(v(x))*d/dx[ln(u(x))*(v(x))] anyone here who have solved anything like this before? Last edited by dorkfork; October 5th, 2016 at 08:26 AM.
 October 5th, 2016, 08:42 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 I assume you mean the function ... $f(x) = (x\sin{x})(x^2+1)^{-3x}$ let $u = x\sin{x}$ $v = (x^2+1)^{-3x}$ start by finding the derivative of $v$ using logarithmic differentiation ... $v = (x^2+1)^{-3x}$ $\ln{v} = -3x\ln(x^2+1)$ $\dfrac{v'}{v} = -3x \cdot \dfrac{2x}{x^2+1} - 3\ln(x^2+1)$ $\dfrac{v'}{v} = -3\bigg[x \cdot \dfrac{2x}{x^2+1} + \ln(x^2+1)\bigg]$ $v' = -3v \bigg[\dfrac{2x^2}{x^2+1} + \ln(x^2+1) \bigg]$ $v' = -3(x^2+1)^{-3x} \bigg[\dfrac{2x^2}{x^2+1} + \ln(x^2+1) \bigg]$ I know, it's a mess. Continuing with $u$ ... $u' = x\cos{x} + \sin{x}$ now you can put it all together using the product rule ... $\dfrac{d}{dx}(u \cdot v) = uv'+u'v$
 October 7th, 2016, 08:37 AM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 seems that you mean for sinx to be an exponent also. let's do logarithmic differentiation for the whole thing ... $f(x) = x^{\sin{x}} \cdot (x^2+1)^{-3x}$ $\ln[f(x)] = \ln \bigg[ x^{\sin{x}} \cdot (x^2+1)^{-3x} \bigg]$ using properties of logs ... $\ln[f(x)] = \ln \left[ x^{\sin{x}}\right] + \ln\left[ (x^2+1)^{-3x} \right]$ $\ln[f(x)] = \sin{x} \cdot \ln{x} - 3x \cdot \ln(x^2+1)$ take the derivative of both sides ... $\dfrac{f'(x)}{f(x)} = \bigg[\dfrac{\sin{x}}{x} + \cos{x} \cdot \ln{x} \bigg] - \bigg[3x \cdot \dfrac{2x}{x^2+1} + 3 \cdot \ln(x^2+1)\bigg]$ $f'(x) = f(x) \bigg(\bigg[\dfrac{\sin{x}}{x} + \cos{x} \cdot \ln{x} \bigg] - 3\bigg[\dfrac{2x^2}{x^2+1} + \ln(x^2+1)\bigg] \bigg)$

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