Using the limit definition, how do you differentiate $f (x) = x^{3} + x$ $f(x)=x^3 + x$ ?

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Answer 1 · 2015-11-14T22:00:49

$\frac{d y}{d x} = 3 x^{2} + 1$ $dy/dx = 3x^2 + 1$

We have

$y = x^{3} + x$ $y = x^3 + x$

So

$dy/dx = lim_(Deltax rarr 0)(Deltay)/(Deltax)$

For ease of typing let's say $Deltax = h$

$dy/dx = lim_(h rarr 0)((x+h)^3 + x + h - x^3 - x)/h$

$dy/dx = lim_(h rarr 0)(x^3 + 3x^2h + 3xh^2 + h^3 + x + h - x^3 - x)/h$

$dy/dx = lim_(h rarr 0)(cancel(x^3) + 3x^2h + 3xh^2 + h^3 cancel(+x) + h cancel(- x^3) cancel(- x))/h$

$dy/dx = lim_(h rarr 0)(3x^2h + 3xh^2 + h^3 + h)/h$

$dy/dx = lim_(h rarr 0)3x^2 + 3xh + h^2 + 1$

$dy/dx = 3x^2 + 1$

Using the limit definition, how do you differentiate f(x)=x^3 + xf(x)=x3+xf(x)=x^3 + x?