My Math Forum Chain Rule, help please

 Calculus Calculus Math Forum

 March 13th, 2009, 06:44 AM #1 Newbie   Joined: Mar 2009 Posts: 1 Thanks: 0 Chain Rule, help please Can someone help solve these equations, using the chain rule? f(x) = x^4 + 4 / x^4 d/dt = 3(t^3 + t^2) ^-2 f(x) = (x-2lnx)^4 Thank you!
 March 13th, 2009, 11:06 AM #2 Senior Member   Joined: Sep 2008 Posts: 116 Thanks: 0 Re: Chain Rule, help please I'm gussing you just want the dirivitives of each? 1) $f(x)= x^4 + \frac{4}{x^4}$ $f'(x) = 4x^3 + 4x^{-4}$ $f'(x) = 4x^3 - 16x^{-5}$ $f'(x) = 4x^3 - \frac{16}{x^5}$ 2) $\frac{d}{dt}= 3(t^3 + t^2)^{-2}$ $\frac{d^2}{dt^2}= -6(t^3 + t^2)^{-3} * \frac{d}{dt}(t^3 + t^2)$ $\frac{d^2}{dt^2}= -6(t^3 + t^2)^{-3} * (3t^2 + 2t)$ $\frac{d^2}{dt^2}= \fract{-6(3t^2 + 2t)}{(t^3 + t^2)^3}$ 3) $f(x)= (x - 2lnx)^4$ $f'(x) = 4(x - 2lnx)^3 * \frac{d}{dt}(x - 2lnx)$ $f'(x) = 4(x - 2lnx)^3 * (1 - 2\frac{1}{x})$ $f'(x) = 4(1 - \frac{2}{x})(x - 2lnx)^3$

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