Question

. Use the definition of the derivative to show that for the function f(x)=3/1x^2−5, the derivative is f(x)=3/2x.

Answer 1

lim h-->0 f(x+h) - f(x) / h

Answer 2

plug and chug

Answer 3

I dont understand this what so ever

Answer 4

you make one big equation...anywhere you see x you replace it with x+h

Answer 5

then you subtract that from the original equation...and most of it should cancel...then you put 0 in for h and there you go

Answer 6

when i plug it in will i get [1/3(x+h)-5-(1/3x ^{2}-5)\]

Answer 7

isn't it 3/x^2-5?

Answer 8

limit the h to 0 and you'll get what you need
f(x) = x^n; f'(x) = nx^n-1
f(x) = constant; f'(x) = 0

Answer 9

tx...he has to use the definition of a derivative

Answer 10

no srry its 1/3

Answer 11

so ti's 1/3x-5?

Answer 12

yes

Answer 13

1/3x^2-5?

Answer 14

okay so then your gonna get (1/3(x+h)^2 -5 ) - (1/3x^2-5)

Answer 15

foil out the (x+h)^2

Answer 16

so you get (1/3(x^2+2h+h^2)-5)- (1/3x^2-5)

Answer 17

(1/3x^2+1/3h^2-5)-(1/3x^2-5)

Answer 18

now distribute the - sign for the 2nd equation

Answer 19

where did 2h come from

Answer 20

(x+h)(x+h) = x^2 +2xh + h^2

Answer 21

thats where im messin up at im just squaring them and now that you said foil im understanding it. imma try and wrk it out

Answer 22

yea....you'll see after you distribute the - sign...alot will cancel

Answer 23

\[\frac{\frac{1}{3}(x+h)^2-5-(\frac{1}{3}x^2-5)}{h}\]
\[=\frac{\frac{1}{3}(x^2+2xh+h^2)-5-\frac{1}{x}x^2+5}{h}\]
\[=\frac{\frac{1}{3}(2xh+h^2)}{h}\]
\[=\frac{\frac{1}{3}h(2x+h)}{h}\]
\[=\frac{1}{3}(2x+h)\] now let h go to 0 get your answer

Answer 24

so wouldnt b -2/3x