The Student Room Group

need help on dy/dx

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??
(edited 12 years ago)

Scroll to see replies

Reply 1
For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate x2x^2, and then substitute your values for x to get the gradient.

Similar thing for x3x^3 except you differentiate x3x^3 instead.
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 x2x^2, and then substitute your values for x to get the gradient.

Similar thing for x3x^3 except you differentiate x3x^3 instead.


how do i do it then
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 x2x^2, and then substitute your values for x to get the gradient.

Similar thing for x3x^3 except you differentiate x3x^3 instead.


ah wait for the first ones is it

x=2 gradient=4

x=6 gradient=12

x=-1 gradient=1?
Reply 4
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.
Reply 5
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.
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
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
Reply 8
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.
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.
Reply 10
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.
Reply 11
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.
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

????
(edited 12 years ago)
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

????
(edited 12 years ago)
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

????
(edited 12 years ago)
When y=an,dydx=nan1y=a^n, \frac{dy}{dx}=na^{n-1} - just apply that to the y=x3y=x^3 and substitute the values of x for the gradients you require.
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??
Reply 17
No.. differentiating x^3 gives 3x^2.. sub in x = 3 and you get 3(3)^2 = 27.
Reply 18
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.
(edited 12 years ago)
Reply 19
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

Quick Reply

Latest