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using the difintion of derivative only find the derivative of
f(x)=x/x^2+1
 11 months ago
 11 months ago
using the difintion of derivative only find the derivative of f(x)=x/x^2+1
 11 months ago
 11 months ago

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SithsAndGigglesBest ResponseYou've already chosen the best response.1
\[\begin{align*}f(x)&=\frac{x}{x^2+1}\\ \\ f'(x)&=\lim_{h\to0}\dfrac{\dfrac{x+h}{(x+h)^2+1}\dfrac{x}{x^2+1}}{h} \end{align*}\] Focusing on the numerator: \[\left(\dfrac{x+h}{(x+h)^2+1}\cdot\dfrac{x^2+1}{x^2+1}\right)\left(\dfrac{x}{x^2+1}\cdot\dfrac{(x+h)^2+1}{(x+h)^2+1}\right)\\ \\ \dfrac{(x+h)(x^2+1)  x((x+h)^2+1)}{((x+h)^2+1)(x^2+1)}\\ \dfrac{x^3 + hx^2+x+h  x^32x^2hxh^2x}{((x+h)^2+1)(x^2+1)}\\ \dfrac{hh^2xhx^2}{((x+h)^2+1)(x^2+1)}\] Now back to the limit: \[\lim_{h\to0}\dfrac{\dfrac{hh^2xhx^2}{((x+h)^2+1)(x^2+1)}}{h}\\ \lim_{h\to0}\dfrac{h\left(\dfrac{1hxx^2}{((x+h)^2+1)(x^2+1)}\right)}{h}\\ \lim_{h\to0}\dfrac{1hxx^2}{((x+h)^2+1)(x^2+1)}\\ \dfrac{1x^2}{(x^2+1)(x^2+1)}\\ \dfrac{1x^2}{(x^2+1)^2}\]
 11 months ago
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