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Now find the slope of the secant lines through (2, 4) and (3,9); (2, 4) and (2.5, (2.5)2); (2, 4) and (2.1, (2.1)2); (2, 4) and (2.01, (2.01)2); (2, 4) and (2.001, (2.001)2); (2, 4) and (2 + (x, (2 + (x)2) ((x represents a small change in x) Find the slope of the tangent line by taking the limit of the last slope in (c) as (x ( 0; recall the slope of the tangent line is the derivative of the function at that point, so you have now found the derivative f ' (2). So our definition of the derivative, f ' (x), of the function f(x) is derived from taking the limit of slopes of secant lines, and is given by:  If this limit exists at a point, then we say f is differentiable at that point. If the limit does not exist at the point, then f is not differentiable at the point. A function that is differentiable everywhere has a derivative for all real numbers. Fact: If a function is differentiable at a point then it is continuous at that point. (So if a function isn’t continuous at a point then it isn't differentiable at that point). Calculate the derivatives of the following functions by definition: f(x) = 3x2 +2x g(x) = 4/x h(x) = sin(x) (4) k(x) = |x|. Note for a piecewise function, if you aren't at a point where the definition changes, then you only use one piece of the function. What I am saying is that there are 3 cases to consider here. x< 0 (then k(x) = -x), x > 0 (where k(x) = x), and then the point in the middle x=0 when you have a left- and right- limit for the derivative just as with checking continuity. This case shows that a function can be continuous at a point yet not differentiable at that point.  EMBED Equation.3  >ЂЃ  $%@A_`tu}~€„…   *+_m) л , -  ‚ р B D E X Y Z [ ^ _ ћїє№є№є№є№є№єъєъє№єъєъєфєлєзєзє№єзєзєвШСвє jEHшџUjЂžt= UVmH jU>*CJjCJUmH jЎ№CJ jD№CJCJH*CJ5CJ 5CJ,>?@с)*+,-./01АБВГdЇЈЉЊЋќљљљїїїђїїїїїїїїђїїїђюююююю„h & F$$>?@с)*+,-./01АБВГdЇЈЉЊЋЌ­ЎЏАБВГДЕ”•–—)*,-( ) л м н ! 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