Correct gradient?

Question

I have a question to do. Simple, but theres something im stuck on.

The curve C has the equation:
Y = 4x^2 + (5-x)/x

The point P lies on C, and has x-coordinate 1.

a) Show that the value of dy/dx at P is 3


Ive tried it, and i keep getting 4...is the gradient really 3? If so, could you show me how?

I have a question to do. Simple, but theres something im stuck on.

The curve C has the equation:
Y = 4x^2 + (5-x)/x

The point P lies on C, and has x-coordinate 1.

a) Show that the value of dy/dx at P is 3


Ive tried it, and i keep getting 4...is the gradient really 3? If so, could you show me how?








EDIT:

I did it this way, but what did i do wrong?

Y=4x^2+(5-x)/x
Y=4x^2 + 5-x(x^-1)

dy/dx= 8x + (-5+x)x^-2
= 8 + (-4)/1
=4

"You must use the product rule for the way you did it, where
f[x]=5-x g[x]= x^-1 f'[x]= -1 g'[x]= -x^-2


Product rule= f'[x]g[x] + f[x]g'[x]"


I dont think we have learnt the product rule yet....:S

Answer 1

Yes, I get 3.

What did you get for the derivative? I rewrote it as:
y = 4x^2 + 5x^-1 - 1
then took the derivative:

dy/dx = 8x - 5x^-2

Put that x-coordinate of 1 into this and indeed you get 8-5 = 3.

that's it! Blessed New Year :)

Answer 2

First we find the derivative:
slope = dy/dx
slope = d/dx(4*x^2 + (5-x)/x)
slope = d/dx(4*x^2 + 5/x - 1)
slope = 8*x - 5/x^2

Then we evaluate for x = 1:
slope = 8*(1) - 5/(1)^2
slope = 8 - 5
slope = 3

Answer 3

?

Answer 4

=8x + ((-x - (5-x)/(x^2))
=8x + (- 5/x^2)
=8x -5/x^2

At point P (where x=1)
dy/dx=8x -5/x^2
dy/dx=8(1) - 5/(1)^2
dy/dx=8-5
dy/dx=3

[Answer: dy/dx at point P is 3]

Use quotient rule on the (5-x)/x part
Quotient rule= f'[x]g[x] - f[x]g'[x]/ (g[x])^2 ,where
f[x]= (5-x)
f'[x] -1
g[x]=x
g'[x]=1


ADDITIONAL DETAILS (WHAT YOU DID):
Y=4x^2+(5-x)/x
Y=4x^2 + 5-x(x^-1)

dy/dx= 8x + (-5+x)x^-2 <-- this is incorrect
You must use the product rule for the way you did it, where
f[x]=5-x g[x]= x^-1 f'[x]= -1 g'[x]= -x^-2
Product rule= f'[x]g[x] + f[x]g'[x]
So we get
dy/dx= 8x + (-x^-1 + (5-x)(-x^-2)
dy/dx= 8(1) + (-(1)^-1 + (5-(1))(-(1)^-2)
dy/dx = 8 + (-1 +4*(-1))
dy/dx = 8 + (-5)
dy/dx = 3