[jetskiwavedunkno]

Find the equations of the normals to the curves at the given point

y=(x+1)^2 at x=0

y=4/x at x=1

answers in form ax+by+c=0

how do i do this

also can some1 check if this is right

find the equations of the tangents to the curves at the given points

y=5x^2+1 at x=2 y=50?

y=3x^2-4x at x=3 y=14??

[mr tim]

For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate , and then substitute your values for x to get the gradient.

Similar thing for except you differentiate instead.

[jetskiwavedunkno]

(Original post by mr tim)
For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate , and then substitute your values for x to get the gradient.

Similar thing for except you differentiate instead.

how do i do it then

[jetskiwavedunkno]

(Original post by mr tim)
For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate , and then substitute your values for x to get the gradient.

Similar thing for except you differentiate instead.

ah wait for the first ones is it

x=2 gradient=4

x=6 gradient=12

x=-1 gradient=1?

[Gemini92]

(Original post by jetskiwavedunkno)
ah wait for the first ones is it

x=2 gradient=4

x=6 gradient=12

x=-1 gradient=1?

The first two are right, but the last one is wrong.

[raheem94]

(Original post by jetskiwavedunkno)
ah wait for the first ones is it

x=2 gradient=4

x=6 gradient=12

x=-1 gradient=1?

Your calculated gradient at x=-1 is wrong, others are correct.

[jetskiwavedunkno]

(Original post by Gemini92)
The first two are right, but the last one is wrong.

x=-1 gradient= -2??? how do i do the x^3 ones

[jetskiwavedunkno]

(Original post by raheem94)
Your calculated gradient at x=-1 is wrong, others are correct.

is the gradient -2? nd how do i do the x^3 ones plz

[raheem94]

(Original post by jetskiwavedunkno)
is the gradient -2? nd how do i do the x^3 ones plz

Yes the gradient is -2.

For the x^3 expression, first differentiate it, than substitute different values of x to find the gradient at different points.

[Contrad!ction.]

(Original post by jetskiwavedunkno)
is the gradient -2? nd how do i do the x^3 ones plz

To differentiate, you'll multiply x by its power, and then subtract 1 from the power.

Eg. differentiating x^6 =6x^5.

Yeah, -2's correct. 2(-1)=-2.

Try using the rule I stated above to differentiate x^3. Actually write what you get in terms of x first so we can see what you're doing.

[Gemini92]

(Original post by jetskiwavedunkno)
x=-1 gradient= -2??? how do i do the x^3 ones

That's right. For the x^3 ones, you have to differentiate x^3 and then substitute each x value into the derivative to get the gradient.

[jordan-s]

For the first one, you need to start by differentiating x^2.
Differentiating it gives you "dy/dy", also known as the gradient function.
The method of differentiating, is "Multiply by the current power, and then decrease the power by 1."

So to differentiate x^2, you multiply by 2, and decrease the power fromk 2 to 1; hence the result is 2x.

Youy would then need to substitute the value of x in to find the gradient at that point.

For x^3, it is a similar process. Start by multiplying by the power, so multiply by 3, and then reduce the power from 3 to 2. Hence the result is 3x^2. Again sub in values of x to find the gradient.

[jetskiwavedunkno]

(Original post by raheem94)
Yes the gradient is -2.

For the x^3 expression, first differentiate it, than substitute different values of x to find the gradient at different points.

is it for x^3

x=3 gradient=27

x=5 gradient=75

x=-3 gradient=27

????

[jetskiwavedunkno]

(Original post by Contrad!ction.)
To differentiate, you'll multiply x by its power, and then subtract 1 from the power.

Eg. differentiating x^6 =6x^5.

Yeah, -2's correct. 2(-1)=-2.

Try using the rule I stated above to differentiate x^3. Actually write what you get in terms of x first so we can see what you're doing.

is it for x^3

x=3 gradient=27

x=5 gradient=75

x=-3 gradient=27

????

[jetskiwavedunkno]

(Original post by Gemini92)
That's right. For the x^3 ones, you have to differentiate x^3 and then substitute each x value into the derivative to get the gradient.

is it for x^3

x=3 gradient=27

x=5 gradient=75

x=-3 gradient=27

????

[Mr Inquisitive]

When - just apply that to the and substitute the values of x for the gradients you require.

[jetskiwavedunkno]

(Original post by jordan-s)
For the first one, you need to start by differentiating x^2.
Differentiating it gives you "dy/dy", also known as the gradient function.
The method of differentiating, is "Multiply by the current power, and then decrease the power by 1."

So to differentiate x^2, you multiply by 2, and decrease the power fromk 2 to 1; hence the result is 2x.

Youy would then need to substitute the value of x in to find the gradient at that point.

For x^3, it is a similar process. Start by multiplying by the power, so multiply by 3, and then reduce the power from 3 to 2. Hence the result is 3x^2. Again sub in values of x to find the gradient.

ahhh crap wait so differentiating 3x is 3x^2 and not 2x^2??

[jordan-s]

No.. differentiating x^3 gives 3x^2.. sub in x = 3 and you get 3(3)^2 = 27.

[raheem94]

(Original post by jetskiwavedunkno)
is it for x^3

x=3 gradient=18

x=5 gradient=50

x=-3 gradient=18

????

Try again, all are wrong.

Differentiating x^3 gives 3x^2.

[Gemini92]

(Original post by jetskiwavedunkno)
ahhh crap wait so differentiating 3x is 3x^2 and not 2x^2??

Yes, the differential of x^3 is 3x^2