Find The Derivate Using The Definition Help

[mop969]

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)

 

[BigGlenntheHeavy]

First one:

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

\(\displaystyle Ergo, \ f \ ' \ (t) \ = \ \lim_{h\to0}\frac{(t+h)+(t+h)^{1/2}-t-t^{1/2}}{h} \ = \ \lim_{h\to0}\frac{(h+(t+h)^{1/2})-t^{1/2}}{h}\)

\(\displaystyle = \ \lim_{h\to0}\frac{[(h+(t+h)^{1/2})-t^{1/2}]}{h}*\frac{[(h+(t+h)^{1/2})+t^{1/2}]}{[(h+(t+h)^{1/2})+t^{1/2}]}\)

\(\displaystyle = \ \lim_{h\to0} \frac{[(h+(t+h)^{1/2})^{2}-t]}{h[h+(t+h)^{1/2}+t^{1/2}]}\)

\(\displaystyle = \ \lim_{h\to0}\frac{h^{2}+2h(t+h)^{1/2}+t+h-t}{h[h+(t+h)^{1/2}+t^{1/2}]}\)

\(\displaystyle = \ \lim_{h\to0}\frac{h^{2}+2h(t+h)^{1/2}+h}{h[h+(t+h)^{1/2}+t^{1/2}]}\)

\(\displaystyle = \ \lim_{h\to0}\frac{h+2(t+h)^{1/2}+1}{[h+(t+h)^{1/2}+t^{1/2}]} \ = \ \frac{2t^{1/2}+1}{2t^{1/2}} \ = \ 1+\frac{1}{2t^{1/2}}. \ QED\)

 

[mop969]

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.

 

[BigGlenntheHeavy]

\(\displaystyle [h+(t+h)^{1/2}]^{2} \ = \ h^{2}+2h(t+h)^{1/2}+t+h\)

 

[mop969]

can you please show me the steps on how you got that? Mainly how did you get 2h?

 

[BigGlenntheHeavy]

\(\displaystyle [h+(t+h)^{1/2}]*[h+(t+h)^{1/2}] \ = \ h^{2}+h(t+h)^{1/2}+h(t+h)^{1/2}+(t+h), \ FOIL \ = \ h^{2}+2h(t+h)^{1/2}+(t+h)\)

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

\(\displaystyle Note: \ (x+y)^{2} \ = \ (x+y)(x+y) \ = \ x^{2}+2xy+y^{2}\)

 

[BigGlenntheHeavy]

Second one:

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

\(\displaystyle f \ ' \ (x) \ = \ \lim_{h\to0}\frac{\frac{x+h-1}{x+h-2}-\frac{x-1}{x-2}}{h}\)

\(\displaystyle = \ \lim_{h\to0}\frac{(x+h-1)(x-2)-(x-1)(x+h-2)}{h(x-2)(x+h-2)}\)

\(\displaystyle = \ \lim_{h\to0}\frac{x^{2}+xh-x-2x-2h+2-x^{2}-xh+2x+x+h-2}{h(x-2)(x+h-2)}\)

\(\displaystyle = \ \lim_{h\to0}\frac{-h}{h(x-2)(x+h-2)} \ = \ \lim_{h\to0}\frac{-1}{(x-2)(x+h-2)} \ = \ \frac{-1}{(x-2)^{2}}, \ QED\)

 

[mop969]

Thank you for all the help problem solved.

 

[mmm4444bot]

… problem solved.

They certainly are!