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The title is pretty self explanitary.
From first principles find f'(x) for f(x)=x.(rootx)
Look forward to your replies
From first principles find f'(x) for f(x)=x.(rootx)
Look forward to your replies
Well first of all that's wrong: f'(x) is not f(a+h)- f(a).NewScientist said:f'(x) = f(a+h) + f(a)
f'(x) = lim (h->0) [(x+h)^3/2 + x^3/2]/h
how do you resolve the (x+h)^3/2)
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This could not possibly have come from a textbook. The definition of the derivative of a function isNewScientist said:f'(x) = f(a+h) + f(a)
Almost. The numerator should be a difference (not a sum).f'(x) = lim (h->0) [(x+h)^3/2 + x^3/2]/h
You rewrite the numerator correctly, and then you multiply the top and bottom of the difference quotient by the conjuagate of the numerator.how do you resolve the (x+h)^3/2)
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Just a small error, it's 3x^{2}h + 3xh^{2} + h^{3} in the numerator (not h^{2}).NewScientist said:[tex]f'(x)=\lim_{h\rightarrow 0}\frac{3x^2h+3h^2x +h^2}{h[(x+h)^\frac{3}{2} + 3^\frac{3}{2}]}[/tex].
Any idea?