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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
chain rule constant rule  ungeheuer  Calculus  1  July 30th, 2013 05:10 PM 
Product rule into chain rule  unwisetome3  Calculus  4  October 19th, 2012 01:21 PM 
Chain Rule..  pomazebog  Calculus  3  February 8th, 2012 01:11 PM 
how do yo solve this one using chain rule with quotient rule  Peter1107  Calculus  1  September 8th, 2011 10:25 AM 
use chain rule only  bgbgbg  Calculus  2  November 6th, 2009 04:56 PM 