Question

find f '(x) of 1/square root of t. give the definition of the derivitative. f '(x)=(f(x+a)-f(x))/(h) I know you find denominator, then conjugate, but now what??

Answer 1

\[\frac{\frac{1}{\sqrt{t+h}} - \frac{1}{\sqrt{t}}}{h}\]
\[= \frac{\sqrt{t} - \sqrt{t+h}}{h \sqrt{t} \sqrt{t+h}}\]
like you said, multiply/divide by conjugate
\[= \frac{t - (t+h)}{h \sqrt{t} \sqrt{t+h} (\sqrt{t} + \sqrt{t +h})}\]
t - t = 0 leaving an h on top .... this cancels with h in denominator
\[= \frac{-1}{\sqrt{t} \sqrt{t+h} (\sqrt{t} + \sqrt{t +h})}\]
plug in h=0 and evaluate
\[= -\frac{1}{2 t \sqrt{t}}\]

Answer 2

thank you thank you!!!!