Question

find f'(a)
f(x)=x^-2

Answer 1

\[f(x)=x^n\]
\[f'(x)=nx^{n-1}\]

Answer 2

or you can do it the long way
\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]

Answer 3

i need to somehow get it into

f(a+h)-f(a)/h

Answer 4

ya were learning the long way

Answer 5

\[f(x)=x^{-2}\]
\[f(y)=y^{-2}\]
\[f(x+100)=(x+100)^{-2}\]
\[f(1)=1^{-2}\]
\[f(x+h)=???\]

Answer 6

(a+h)^-2 - (a)^-2
---------------
h

Answer 7

im blanking on negative exponents, let me check my notes

Answer 8

right now
lets just ignore the h for now
we'll get back to that
\[(a+h)^{-2} - (a)^{-2}=\frac{1}{(a+h)^{2}} - \frac{1}{(a)^2}\]

Answer 9

now just simplify that equation, then put it all over h

Answer 10

ohhhhh your the man/woman

Answer 11

i prefer idiot

Answer 12

lol

Answer 13

thanks a lot, that really made things clear

Answer 14

hey idiot, i got a weird answer after working through it

Answer 15

f(x) = x^-2
let f(x) = y

y=x^-2
y=1/x^2
x^2=1/y
x=\[\sqrt{(1/y)}\]
therefore

f'(x)=\[\sqrt{(1/x)}\]

Answer 16

\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[f(x)=x^{-2}\]
\[\lim_{h \rightarrow 0}\frac{(x+h)^{-2}-x^{-2}}{h}\]
\[\lim_{h \rightarrow 0}\frac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}\]
\[\lim_{h \rightarrow 0}\frac{\frac{1}{x^2+2xh+h^2}-\frac{1}{x^{2}}}{h}\]
\[\lim_{h \rightarrow 0}\frac{\frac{x^2}{x^2(x^2+2xh+h^2)}-\frac{x^2+2xh+h^2}{x^{2}(x^2+2xh+h^2)}}{h}\]
\[\lim_{h \rightarrow 0}\frac{\frac{-2xh-h^2}{x^{2}(x^2+2xh+h^2)}}{h}\]
\[\lim_{h \rightarrow 0}\frac{-2x-h}{x^{2}(x^2+2xh+h^2)}\]
\[\frac{-2x}{x^2*x^2}\]
\[\frac{-2x}{x^4}\]
\[\frac{-2}{x^3}\]
\[f'(x)=-2x^{-3}\]

Answer 17

you can only cross multiply only if the 2 fractions are equal to each other

in this case you can only find the GCD or LCD whichever it is
and then combine it, kind of like cross multiplying