Find The Derivate Using The Definition Help

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mop969

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I know what the definition to find the derivative is but I get stuck when solving this problem. Please show the steps to find the derivative of this problem using the definition. The problem is:

f(t) = t + ?t


Also I am stuck with this problem too it seems like when I use the definition everything is canceling leaving me with one which is not right:


y = (x - 1)/(x - 2)
 
First one:

f(t) = t+t1/2, f(t+h) = (t+h)+(t+h)1/2\displaystyle f(t) \ = \ t+t^{1/2}, \ f(t+h) \ = \ (t+h)+(t+h)^{1/2}

Ergo, f ′ (t) = lim⁡h→0(t+h)+(t+h)1/2−t−t1/2h = lim⁡h→0(h+(t+h)1/2)−t1/2h\displaystyle Ergo, \ f \ ' \ (t) \ = \ \lim_{h\to0}\frac{(t+h)+(t+h)^{1/2}-t-t^{1/2}}{h} \ = \ \lim_{h\to0}\frac{(h+(t+h)^{1/2})-t^{1/2}}{h}

= lim⁡h→0[(h+(t+h)1/2)−t1/2]h∗[(h+(t+h)1/2)+t1/2][(h+(t+h)1/2)+t1/2]\displaystyle = \ \lim_{h\to0}\frac{[(h+(t+h)^{1/2})-t^{1/2}]}{h}*\frac{[(h+(t+h)^{1/2})+t^{1/2}]}{[(h+(t+h)^{1/2})+t^{1/2}]}

= lim⁡h→0[(h+(t+h)1/2)2−t]h[h+(t+h)1/2+t1/2]\displaystyle = \ \lim_{h\to0} \frac{[(h+(t+h)^{1/2})^{2}-t]}{h[h+(t+h)^{1/2}+t^{1/2}]}

= lim⁡h→0h2+2h(t+h)1/2+t+h−th[h+(t+h)1/2+t1/2]\displaystyle = \ \lim_{h\to0}\frac{h^{2}+2h(t+h)^{1/2}+t+h-t}{h[h+(t+h)^{1/2}+t^{1/2}]}

= lim⁡h→0h2+2h(t+h)1/2+hh[h+(t+h)1/2+t1/2]\displaystyle = \ \lim_{h\to0}\frac{h^{2}+2h(t+h)^{1/2}+h}{h[h+(t+h)^{1/2}+t^{1/2}]}

= lim⁡h→0h+2(t+h)1/2+1[h+(t+h)1/2+t1/2] = 2t1/2+12t1/2 = 1+12t1/2. QED\displaystyle = \ \lim_{h\to0}\frac{h+2(t+h)^{1/2}+1}{[h+(t+h)^{1/2}+t^{1/2}]} \ = \ \frac{2t^{1/2}+1}{2t^{1/2}} \ = \ 1+\frac{1}{2t^{1/2}}. \ QED
 
Thank You. But I am lost with the 5th row down how did you get h^2 plus 2h etc. from the previous line? Thanks.
 
[h+(t+h)1/2]2 = h2+2h(t+h)1/2+t+h\displaystyle [h+(t+h)^{1/2}]^{2} \ = \ h^{2}+2h(t+h)^{1/2}+t+h
 
can you please show me the steps on how you got that? Mainly how did you get 2h?
 
[h+(t+h)1/2]∗[h+(t+h)1/2] = h2+h(t+h)1/2+h(t+h)1/2+(t+h), FOIL = h2+2h(t+h)1/2+(t+h)\displaystyle [h+(t+h)^{1/2}]*[h+(t+h)^{1/2}] \ = \ h^{2}+h(t+h)^{1/2}+h(t+h)^{1/2}+(t+h), \ FOIL \ = \ h^{2}+2h(t+h)^{1/2}+(t+h)

(t+h)1/2∗(t+h)1/2 = t+h.\displaystyle (t+h)^{1/2}* (t+h)^{1/2} \ = \ t+h.

Note: (x+y)2 = (x+y)(x+y) = x2+2xy+y2\displaystyle Note: \ (x+y)^{2} \ = \ (x+y)(x+y) \ = \ x^{2}+2xy+y^{2}
 
Second one:

f(x) = x−1x−2\displaystyle f(x) \ = \ \frac{x-1}{x-2}

f ′ (x) = lim⁡h→0x+h−1x+h−2−x−1x−2h\displaystyle f \ ' \ (x) \ = \ \lim_{h\to0}\frac{\frac{x+h-1}{x+h-2}-\frac{x-1}{x-2}}{h}

= lim⁡h→0(x+h−1)(x−2)−(x−1)(x+h−2)h(x−2)(x+h−2)\displaystyle = \ \lim_{h\to0}\frac{(x+h-1)(x-2)-(x-1)(x+h-2)}{h(x-2)(x+h-2)}

= lim⁡h→0x2+xh−x−2x−2h+2−x2−xh+2x+x+h−2h(x−2)(x+h−2)\displaystyle = \ \lim_{h\to0}\frac{x^{2}+xh-x-2x-2h+2-x^{2}-xh+2x+x+h-2}{h(x-2)(x+h-2)}

= lim⁡h→0−hh(x−2)(x+h−2) = lim⁡h→0−1(x−2)(x+h−2) = −1(x−2)2, QED\displaystyle = \ \lim_{h\to0}\frac{-h}{h(x-2)(x+h-2)} \ = \ \lim_{h\to0}\frac{-1}{(x-2)(x+h-2)} \ = \ \frac{-1}{(x-2)^{2}}, \ QED
 
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