Differentiation

If the tangent at x=p has gradient g (i.e. here g = p^3), then it must have equation

y = gx + c for some constant c.

You choose c so that gx+c passes through the point on the curve where x = p.

Reply 7

OP

Original post by RDKGames

(-1)^4 \not = -1

Also as noted above, post the the full question.

I did below.

Original post by TheMightyAugur

Sub in P=-1 for your gradient and solve for M

Then sub X = -1 and Y = 1 and M into the equation

Y - Y1 = M (X - X1)

Y - 1 = -4 (X + 1)

Y - 1 = -4X - 4

Y + 4X + 3 = 0

Posted from TSR Mobile

So x = -1 also means p = -1??

Original post by ckfeister

So x = -1 also means p = -1??

Yes. That would be the point P if you refer to the context of the question. You have derived the gradient at this point from first principles.

Reply 9

OP

Original post by RDKGames

Yes. That would be the point P if you refer to the context of the question. You have derived the gradient at this point from first principles.

Original post by DFranklin

If the tangent at x=p has gradient g (i.e. here g = p^3), then it must have equation

y = gx + c for some constant c.

You choose c so that gx+c passes through the point on the curve where x = p.

Original post by notnek

Can you please post the full question?

Original post by TheMightyAugur

Sub in P=-1 for your gradient and solve for M

Then sub X = -1 and Y = 1 and M into the equation

Y - Y1 = M (X - X1)

Y - 1 = -4 (X + 1)

Y - 1 = -4X - 4

Y + 4X + 3 = 0

Posted from TSR Mobile

4. A and B lie on the curve y = 1/x at the points where x = -2 and x = -2 + h, respectively.

(i) Find the gradient of the line AB.

y = \frac{1}{x}

A(-2, - \frac{1}{2})

B(-2+h, - \frac{1}{-2+h})

m = \frac{- \frac{1}{2} + \frac{1}{-2+h}}{h}

How do I finish this off?

Never seen questions like this before...

(edited 7 years ago)

Original post by ckfeister

4. A and B lie on the curve y = 1/x at the points where x = -2 and x = -2 + h, respectively.

(i) Find the gradient of the line AB.

y = \frac{1}{x}

A(-2, -1/2)

B(-2+h, - \frac{1}{-2+h})

m = \frac{- \frac{1}{2} + \frac{1}{-2+h}}{-2+2+h} [br]m = \frac{- \frac{1}{2} + \frac{1}{-2+h}}{h}

How do I finish this off?

Never seen questions like this before...

Just simplify and leave your answer in terms of h.

Reply 11

Rather than go straight for "what is m", you may want to calculate separately

\Delta x

= the change in x (so -2 - (-2-h)) here.

\Delta y

= the change in y (so

\displaystyle \frac{-1}{2} - \frac{-1}{2-h}

) here.

Get both of these down to single fraction expressions and then combine to find

m = \dfrac{\Delta y}{\Delta x}

.

It's not really any different from what you're doing but the individual expressions are simpler and you're a lot less likely to get confused than if you have expressions like

\dfrac{\frac{-1}{2} + \frac{1}{-2+h}}{-2+2+h}

(it's easier to LaTeX, too :smile:

)

Reply 12

OP

Original post by RDKGames

Just simplify and leave your answer in terms of h.

I've got to

m = -h/2 + h/-2+h

What do I do from here??
Answer is
(i) 1/ 2(h - 2)

Reply 13

Original post by ckfeister

I've got to

m = -h/2 + h/-2+h

\dfrac{1}{A} - \dfrac{1}{B} = \dfrac{B-A}{AB}

Original post by ckfeister

I've got to

m = -h/2 + h/-2+h

What do I do from here??

That's not right... where did the h come from in the numerator?

Reply 15

OP

Original post by RDKGames

That's not right... where did the h come from in the numerator?

\frac {- \frac {1}{2} + \frac {1}{h-2}}{h}

I've never seen any questions or examples on these, idk what I'm doing but I did ..

- \frac{1}{2}{h} + \frac {1}{h-2}{h}

m = - \frac{h}{2} + \frac{2}{h-2}

(edited 7 years ago)

Reply 16

Original post by ckfeister

\frac {- \frac {1}{2} + \frac {1}{h-2}}{h}

I've never seen any questions or examples on these, idk what I'm doing but I did ..

- \frac{1}{2}{h} + \frac {1}{h-2}{h}

If the first line you were dividing by h, and it has mysteriously changed into you multiplying by h. This is wrong.

Reply 17

OP

Original post by DFranklin

If the first line you were dividing by h, and it has mysteriously changed into you multiplying by h. This is wrong.

a/b/c = ac/b correct? as a/b / 1/c = a/b * c/1 = a/b *c/1 = ac/b

Original post by ckfeister

\frac {- \frac {1}{2} + \frac {1}{h-2}}{h}

I've never seen any questions or examples on these, idk what I'm doing but I did ..

- \frac{1}{2}{h} + \frac {1}{h-2}{h}

m = - \frac{h}{2} + \frac{2}{h-2}

Yeah those two are not equal to one another. If you multiply the numerator of the main fraction by h then you must multiply the denominator of the main fraction by h, thus giving you

h^2

on the main denominator which not what you want. Also it should be

-h

on your main denominator by the looks of it anyway.

To carry on correctly, simplify

-\frac{1}{2}+\frac{1}{h-2}

first.

(edited 7 years ago)

Reply 19