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Differentiation

Find the gradient of the curve with equation y=f(x) at the point A where:

f(x)=x^3-3x+2 and A is at (-1,4)

This is what I've done so far but I think it's wrong.

Eqn2.gif
Original post by zed963
Find the gradient of the curve with equation y=f(x) at the point A where:

f(x)=x^3-3x+2 and A is at (-1,4)

This is what I've done so far but I think it's wrong.

Eqn2.gif


You have gone wrong twice, First with your differentiation. and why have you differentiated twice !!
ddx(3x)=3 \frac{d}{dx}(3x)=3 so dydx=3x2−3 \frac{dy}{dx}=3x^2-3
x=−1  ⟹  3x2=3x=-1 \implies 3x^2=3 so final answer should be 0
Reply 2
Avatar for krisshP
krisshP
14
Original post by zed963
Find the gradient of the curve with equation y=f(x) at the point A where:

f(x)=x^3-3x+2 and A is at (-1,4)

This is what I've done so far but I think it's wrong.

Eqn2.gif

Like the above poster ^ said, EVERY term gets differentiated. So -3x and 2 also get differentiated. Finding out dy/dx means finding an equation that allows you to work out the curve's gradient at a point, which is what you need to do just once. But differentiating a further time means you finding an equation that allows you to determine at a point the rate of change of the gradient. This is not what you want, so you shouldn't have differentiated the extra time.:smile:
Reply 3
Avatar for krisshP
krisshP
14
Also I forgot to mention that if you have a curve's equation and you differentiate it, you get dy/dx. Differentiating one more time gives you d2ydx2\frac{d^2 y}{dx^2}. Differentiating again gives d3ydx3\frac{d^3 y}{dx^3}. Differentiating another time gives you d4ydx4\frac{d^4 y}{dx^4} etc...
(edited 10 years ago)
Reply 4
Avatar for zed963
zed963
OP
16
Original post by krisshP
Also I forgot to mention that if you have a curve's equation and you differentiate it, you get dy/dx. Differentiating one more time gives you d2ydx2\frac{d^2 y}{dx^2}. Differentiating again gives d3ydx3\frac{d^3 y}{dx^3}. Differentiating another time gives you d4ydx4\frac{d^4 y}{dx^4} etc...



Original post by brianeverit
You have gone wrong twice, First with your differentiation. and why have you differentiated twice !!
ddx(3x)=3 \frac{d}{dx}(3x)=3 so dydx=3x2−3 \frac{dy}{dx}=3x^2-3
x=−1  ⟹  3x2=3x=-1 \implies 3x^2=3 so final answer should be 0


Oh okay,

What I've done now.

 f(x)=x3−3x+2\ f(x)=x^3-3x+2

dydx=3x2−3+2=3x2−1\frac{dy}{dx}=3x^2-3+2=3x^2-1

 3(−1)2−1=2\ 3(-1)^2-1=2
(edited 10 years ago)
Reply 5
Avatar for krisshP
krisshP
14
Original post by zed963
Oh okay,

What I've done now.

 f(x)=x3−3x+2\ f(x)=x^3-3x+2

dydx=3x2−3+2=3x2−1\frac{dy}{dx}=3x^2-3+2=3x^2-1

 3(−1)2−1=2\ 3(-1)^2-1=2


NO!

The 2 is differentiated to 0. Look

2
=2x^0
=0 X 2x^-1

0 x anything=0

so differentiating 2 gives 0.
Reply 6
Avatar for zed963
zed963
OP
16
Original post by krisshP
NO!

The 2 is differentiated to 0. Look

2
=2x^0
=0 X 2x^-1

0 x anything=0

so differentiating 2 gives 0.


Oh...

Yes that's true.

 3x2−3\ 3x^2-3

 3(−1)2−3=0\ 3(-1)^2-3=0
Reply 7
Avatar for zed963
zed963
OP
16
Original post by krisshP
NO!

The 2 is differentiated to 0. Look

2
=2x^0
=0 X 2x^-1

0 x anything=0

so differentiating 2 gives 0.


The two is on its own, why have you included an x?
Reply 8
Avatar for krisshP
krisshP
14
Original post by zed963
The two is on its own, why have you included an x?


Because it allows us to see clearly and easily how we can differentiate it.

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