Question 1190590
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f(x) = x^2
f(x+h) = (x+h)^2 ... every x is replaced with x+h
f(x+h) = x^2+2xh+h^2 ... expand with the FOIL rule


Apply the difference quotient
{{{(f(x+h)-f(x))/h = ((x^2+2xh+h^2)-(x^2))/h}}}


{{{(f(x+h)-f(x))/h = (x^2+2xh+h^2-x^2)/h}}}


{{{(f(x+h)-f(x))/h = (2xh+h^2)/h}}} The x^2 terms cancel out, leaving nothing but terms with h in the numerator


{{{(f(x+h)-f(x))/h = (h(2x+h))/h}}} That way we can factor out h


{{{(f(x+h)-f(x))/h = (highlight(h)(2x+h))/(highlight(h))}}}


{{{(f(x+h)-f(x))/h = (cross(h)(2x+h))/(cross(h))}}} The h's cancel


{{{(f(x+h)-f(x))/h = 2x+h}}}
Now let h approach 0. It won't actually reach 0 itself, but only get closer and closer. This is the limit process in action. Effectively we replace h with 0 (refer to the substitution limit theorem) and we go from 2x+h to 2x+0 or simply 2x.


In short, the derivative of f(x) = x^2 is f ' (x) = 2x.
You'll learn a shortcut (if you haven't already) that's useful and that's the power rule. That will mean you can quickly differentiate any polynomial like this without having to resort to these limits. However it's good to know how all this works out.


Plug x = 3 into the derivative function to find the tangent slope
f ' (x) = 2x
f ' (3) = 2*3
f ' (3) = 6
The tangent slope is m = 6


Answer: <font color=red>6</font>
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