Question

How to find the derivative of f(t)=t^2-6t-5

Answer 1

have you been taught the power rule?
\[y = x^n\implies\frac{dy}{dx}=nx^{n-1}\]?

Answer 2

n can be rational...

Answer 3

Yes but for this we are not supposed to use it

Answer 4

We are supposed to use the normal deriva/ formauls

Answer 5

then you need to use
\[\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\]

Answer 6

So do I substitute the equation in for x?

Answer 7

im not going to write the limit over and over but
\[\frac{f(t+h)-f(t)}{h}=\frac{(t+h)^2-6(t+h)-5}{h}\]

Answer 8

can you simplify the numerator of the one on the left?

Answer 9

err right*

Answer 10

Yea okay thank you so much!!

Answer 11

let me know what you get at the end so I know its right

Answer 12

is it 2t+h-5

Answer 13

should be 2t-6

Answer 14

after you run the limit

Answer 15

How do you find the limit to sub/ into it

Answer 16

we are letting h go to 0 so just make h 0

Answer 17

but thats not the problem

Answer 18

I can't find where I made a mistake

Answer 19

sorry I think I messed up the code showing you the first time....

Answer 20

nvm I found it :( and it was a big wooping stupid mistake, thank you thought!

Answer 21

\[\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2-6(x+h)-5 - (x^2-6x-5)}{h}\\=\frac{x^2+2xh+h^2-6x-6h-5-x^2+6x+5}{h}=\frac{2xh+h^2-6h}{h}=\\2x+h-6\]we let h go to 0
\[=2x+6\]

Answer 22

the first think I showed you was wrong...I left off the -f(x) ...woops

Answer 23

the first thing I showed you was missing the red\[\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2-6(x+h)-5 \color{red}{- (x^2-6x-5)}}{h}\]

Answer 24

Okay cool thank you so much I finally get how to do d/dx