Value of a Dirivative at the point.

[cyclops101]

I am having trouble with this group of problems, if you could help me with just one i could figure out the rest.

Find the vaule of the derivative of the funcion at the indicated point.

y=(2x+1)^2 point0,1)

 

[Gene]

u=2x + 1
d(u²)=2u*du

 

[cyclops101]

uhh... think im more confused then before can u explain that please

 

[Gene]

u=2x+1
d(y) = d(u²)= 2udu =
2(2x+1)*d(2x+1) =
2(2x+1)(2)dx =
(8x+4)dx
dy/dx = 8x+4
Then you plug in x=0

You should show how much you got so we can better help.

 

[pav]

(2x+1)^2

dy/dx= dy/dx(outside) * dy/dx(inside)


dy/dx= 2(2x+1)^1*(2x^(1-1))

dy/dx=(8x+4)*2

dy/dx= 16x+8

at (1,0)
when x=1
dy/dx= 16*(1)+8

 

[pav]

whoops got the point mixed up but just change em
yeah!

 

[Gene]

PAV, check your math. You got an extra 2 someplace. 8x+4 is the final answer. Trust me.
--------------------
Gene

 

[pav]

yeah thanks Gene i did it too fast

2(2x+1)*(2x^(1-1)+1^(1-1))

2(2x+2)*2

4(2x+2)

dy/dx=8x+4