#need help plz
172 messages · Page 1 of 1 (latest)
Hi thanks for replying
um
are the answers -10
and -1 by any chance?
i used the product rule for a
and i multiplyrd the two functions
which gave me -10
and then for b i got -1
used the chain rule
is that right?
Can I see your workings for a.)
They’re asking for u’(-3)
So by product rule:
$u’(x)=f’(x)g(x)+f(x)g’(x)$
denzio321
What is f’(-3)
-7?
Use rise/run formula
rise run formula
The gradient of a line formula
Well in the first place the line is going upwards
So it can’t even be a negative gradient
y2-y1/x2-x1
What did you find f’(-3) to be first
2?
So you picked the points (-2,1) and (-4,-3) right
i used 1-(-3)/-2-(-4)
but u said use thge ppoints only on the line
so it would have to be this what u said
The points u picked aren’t even on the line
Using these points
$m=\frac{0-(-1)}{1-(-3)}=\frac{1}{4}$
Try using these points drawn in black @elder birch
hii sorrry
my wifi went out
im back
im still confused
i tried the points you gave me but
idk
like i got this question right
but i cant get the graph question right
i dont know if im overthinking it, or if i dont understand graphs
lol
Using these points
$m=\frac{0-(-1)}{1-(-3)}=\frac{1}{4}$
denzio321
The points here
Try for the g(x)
okay
What's g'(-3)
Like this
how do i draw?
Use remix
i dont know how to do that sry
Just tell me the points you used
okay
In (x, y) form
(-4,-3) (-2,-4)
Are those points on the line?
Yeps
Yes
So now just find u'(-3)
How'd you work that out?
i did u'(-3) =f'(-3)
No the formula says
u'(x) =f'(x)g(x) +f(x) g'(x)
Where are you getting
u'(-3)=f'(-3) from
Use the graph cuh
What's f(-3)
I'm not talking about f'(-3),that's the derivative
-1?
1
Yep so what is u'(-3)
finallyyy lol
sorry for bothering you i was just confused now i understand thanks
so the answers for 1a is 13/4, now how would i go about solving b
would i do the same thing
just instead of -3
this time 4?
Just wait