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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 ... $\dfrac{f'(x)}{f(x)} = \bigg[\dfrac{\sin{x}}{x} + \cos{x} \cdot \ln{x} \bigg] - \bigg[3x \cdot \dfrac{2x}{x^2+1} + 3 \cdot \ln(x^2+1)\bigg]$ $f'(x) = f(x) \bigg(\bigg[\dfrac{\sin{x}}{x} + \cos{x} \cdot \ln{x} \bigg] - 3\bigg[\dfrac{2x^2}{x^2+1} + \ln(x^2+1)\bigg] \bigg)$ Tags chain, combinding, derivative, power, rule Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ungeheuer Calculus 1 July 30th, 2013 05:10 PM unwisetome3 Calculus 4 October 19th, 2012 01:21 PM pomazebog Calculus 3 February 8th, 2012 01:11 PM Peter1107 Calculus 1 September 8th, 2011 10:25 AM bgbgbg Calculus 2 November 6th, 2009 04:56 PM

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