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# need help on dy/dx

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1. 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??
2. 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. (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. (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. (Original post by jetskiwavedunkno)
ah wait for the first ones is it

The first two are right, but the last one is wrong.
6. (Original post by jetskiwavedunkno)
ah wait for the first ones is it

7. (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. (Original post by raheem94)
is the gradient -2? nd how do i do the x^3 ones plz
9. (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. (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. (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. 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. (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

????
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

????
15. (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

????
16. When - just apply that to the and substitute the values of x for the gradients you require.
17. (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. No.. differentiating x^3 gives 3x^2.. sub in x = 3 and you get 3(3)^2 = 27.
19. (Original post by jetskiwavedunkno)
is it for x^3

????
Try again, all are wrong.

Differentiating x^3 gives 3x^2.
20. (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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