bimbo36 Posted November 13, 2011 Share Posted November 13, 2011 derivative (outer function ) (whole terms) * derivative (inner function) ? Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted November 13, 2011 Share Posted November 13, 2011 Yes. If it helps, you can rewrite it this way: [math](\sin (x^3))^4[/math] Link to comment Share on other sites More sharing options...
bimbo36 Posted November 14, 2011 Author Share Posted November 14, 2011 i have sort of figured out chain rule for two composite functions .. and thank god it works .. sin x xrise 4 xrise 3 so this is three composite functions .. i cant get the answer the way i am working with .. i wish i had more examples to work out, of functions including only "two composite functions" i will deal with three , later ... Link to comment Share on other sites More sharing options...
bimbo36 Posted November 15, 2011 Author Share Posted November 15, 2011 d/dx sin = cosx d/dx cosx = - sin x d/dx tanx = sec2x then chain rule ... composite two functions and composite three functions ... d/dx f(x)og(x) Df(x)og(x) * Dg(x) ------------------------------------------------------------------------------ d/dx f(x)og(x)oh(x) Df(x)og(x)oh(x) * Dg(x)oh(x) *D h(x) Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted November 15, 2011 Share Posted November 15, 2011 Looks like you're on the right track. Here's a simple example: [math]\frac{d}{dx} ((x^5)^4)^3 = 3((x^5)^4)^2 \cdot \frac{d}{dx} (x^5)^4 = 3((x^5)^4)^2 \cdot 4 (x^5)^3 \cdot \frac{d}{dx} x^5[/math] Link to comment Share on other sites More sharing options...
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