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\newline \dfrac{\mbox{d}}{\mbox{d}x} \mbox{f}(\mbox{g}(x)} = \mbox{f}'(\mbox{g}(x))\mbox{g}'(x)
\mbox{f}(x) = \arctan x,\ \mbox{g}(x) = \dfrac{2+x}{1-2x}
The chain rule gives you a general form f(g(x)). Not everything is anywhere near as simple as the general form you gave, and this example in particular isn't.
\newline \dfrac{\mbox{d}}{\mbox{d}x} \mbox{f}(\mbox{g}(x)} = \mbox{f}'(\mbox{g}(x))\mbox{g}'(x)
\mbox{f}(x) = \arctan x,\ \mbox{g}(x) = \dfrac{2+x}{1-2x}
i dno how to do f'(g(x)) > <
Have you done C3? It's just functions stuff.
If you had, say,
\mbox{f}(x) = x^2 + 3x,\ \mbox{g}(x) = \sqrt{x} + 4
\mbox{f}'(x) = 2x + 3
\mbox{f}'(\mbox{g}(x)) = 2\mbox{g}(x) + 3 = 2(\sqrt{x} + 4) + 3 = 2\sqrt{x} + 11
\mbox{g}'(x) = \dfrac{1}{2\sqrt{x}}
\dfrac{d}{dx} \mbox{f}(\mbox{g}(x)) = \dfrac{2\sqrt{x} + 11}{2\sqrt{x}}
You do, you just don't know you do.
Instead of "g(x)", I'm going to write "g"; now, f(g(x)) = f(g), which is just arctan(g). So f'(g(x)) = f'(g) = df/dg. You know how to differentiate arctan(g) with respect to g, because it's exactly the same way you'd differentiate arctan(x) with respect to x, just with a different letter...
If you had, say,
\mbox{f}(x) = x^2 + 3x,\ \mbox{g}(x) = \sqrt{x} + 4
\mbox{f}'(x) = 2x + 3
\mbox{f}'(\mbox{g}(x)) = 2\mbox{g}(x) + 3 = 2(\sqrt{x} + 4) + 3 = 2\sqrt{x} + 11
\mbox{g}'(x) = \dfrac{1}{2\sqrt{x}}
\dfrac{d}{dx} \mbox{f}(\mbox{g}(x)) = \dfrac{2\sqrt{x} + 11}{2\sqrt{x}}
many thanks i understand now
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