Question

f(x) =1/(1-x)^2 obtain f ' (x) using the defining formula

Answer 1

You could use the quotient rule, but i think the chain rule is easier... convert the fraction to a negative power and use the chain rule. \[ \frac{1}{(1-x)^2}=(1-x)^{-2}\]
Chain rule:
\[\frac{d}{dx}f(x)= -2(1-x)^{-3}(-1)=2(1-x)^{-2}=\frac{2}{(1-x)^2}\]

Answer 2

no its the one with (f(x+h)-f(x))/h

Answer 3

oops i changed the 3 into a 2 half-way through

Answer 4

ooh the long way

Answer 5

Lol

Answer 6

\[\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}\\
=\lim_{h\rightarrow 0}\large\frac{\frac{1}{(1-(x+h))^2}-\frac{1}{(1-x)^2}}{h}\\
\, \\

=\lim_{h\rightarrow 0}\large \frac{\frac{1}{((1-x)-h)^2}-\frac{1}{(1-x)^2}}{h}\\
\, \\
=\lim_{h\rightarrow 0}\large \frac{\frac{(1-x)^2}{(1-x)^2((1-x)-h)^2}-\frac{((1-x)-h)^2}{(1-x)^2((1-x)-h)^2}}{h}\\
\, \\
=\lim_{h\rightarrow 0} \large \frac{\frac{(1-x)^2-((1-x)-h)^2}{(1-x)^2((1-x)-h)^2}}{h}\\
\, \\
=\lim_{h\rightarrow 0}\frac{(1-x)^2-((1-x)-h)^2}{(1-x)^2((1-x)-h)^2h}\\ \, \\
=\lim_{h\rightarrow 0} \frac{(1-x)^2-[(1-x)^2-2h(1-x)+h^2]}{(1-x)^2[(1-x)^2-2h(1-x)+h^2]h}\\ \, \\
=\lim_{h\rightarrow 0} \frac{2h(1-x)-h^2}{(1-x)^2h[(1-x)^2-2h(1-x)+h^2]} \\ \, \\
=\lim_{h\rightarrow 0} \frac{\cancel{h}[2(1-x)-h]}{(1-x)^2\cancel{h}[(1-x)^2 -2h(1-x)+h^2]}\], plug in \(h=0\):
\[ =\frac{2(1-x)}{(1-x)^4}\\
= \frac{2}{(1-x)^3}\]

Answer 7

:)

Answer 8

same answer as before so it's good

Answer 9

Thanks!!

Answer 10

yw