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# need help on dy/dx

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1. 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

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??
Last edited by jetskiwavedunkno; 23-02-2012 at 21:13.
2. Re: need help on dy/dx
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.
3. Re: need help on dy/dx
(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
4. Re: need help on dy/dx
(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

5. Re: need help on dy/dx
(Original post by jetskiwavedunkno)
ah wait for the first ones is it

The first two are right, but the last one is wrong.
6. Re: need help on dy/dx
(Original post by jetskiwavedunkno)
ah wait for the first ones is it

7. Re: need help on dy/dx
(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
8. Re: need help on dy/dx
(Original post by raheem94)
is the gradient -2? nd how do i do the x^3 ones plz
9. Re: need help on dy/dx
(Original post by jetskiwavedunkno)
is the gradient -2? nd how do i do the x^3 ones plz

For the x^3 expression, first differentiate it, than substitute different values of x to find the gradient at different points.
10. Re: need help on dy/dx
(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.
11. Re: need help on dy/dx
(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.
12. Re: need help on dy/dx
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.
13. Re: need help on dy/dx
(Original post by raheem94)

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

????
Last edited by jetskiwavedunkno; 23-02-2012 at 20:41.
14. Re: need help on dy/dx
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

????
Last edited by jetskiwavedunkno; 23-02-2012 at 20:42.
15. Re: need help on dy/dx
(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

????
Last edited by jetskiwavedunkno; 23-02-2012 at 20:42.
16. Re: need help on dy/dx
When - just apply that to the and substitute the values of x for the gradients you require.
17. Re: need help on dy/dx
(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??
18. Re: need help on dy/dx
No.. differentiating x^3 gives 3x^2.. sub in x = 3 and you get 3(3)^2 = 27.
19. Re: need help on dy/dx
(Original post by jetskiwavedunkno)
is it for x^3

????
Try again, all are wrong.

Differentiating x^3 gives 3x^2.
Last edited by raheem94; 23-02-2012 at 20:55.
20. Re: need help on dy/dx
(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

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Last updated: February 23, 2012
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