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# need help on dy/dx Watch

1. (Original post by Mr Inquisitive)
When - just apply that to the and substitute the values of x for the gradients you require.
is it x=3 graient =27

2. (Original post by jetskiwavedunkno)
is it x=3 graient =27

Those are all right.
3. (Original post by raheem94)
Only the gradient at x=3 is correct, others are wrong, try again.

How did you differentiated x^3? maybe you are making a mistake there.
you are wrong
4. (Original post by Gemini92)
Those are all right.
awww thank you! xx
5. (Original post by jetskiwavedunkno)
is it x=3 graient =27

Yes, they're correct.
6. (Original post by raheem94)
Only the gradient at x=3 is correct, others are wrong, try again.
I'm afraid the gradient at 3 is also incorrect?

3x^2 when x = 3 gives 3 multiplied by 3^2, which is 3 x 9 = 27..

Ooops, too late! You've got it now anyway.
7. (Original post by Mr Inquisitive)
Yes, they're correct.
for each curve find the gradient at the given point.

how do i do this
8. (Original post by jetskiwavedunkno)
for each curve find the gradient at the given point.

how do i do this
Exactly the same way as the others, differentiate the function and sub in the x value. (Remember that constants differentiate to 0)
9. (Original post by jetskiwavedunkno)
for each curve find the gradient at the given point.

how do i do this
the same way as you did before differentiate each equation, except this time you only need to sub in one x value for each equation, which is 1
10. (Original post by Gemini92)
Exactly the same way as the others, differentiate the function and sub in the x value. (Remember that constants differentiate to 0)

11. (Original post by mr tim)
the same way as you did before differentiate each equation, except this time you only need to sub in one x value for each equation, which is 1
for each curve find the gradient at the given point.

12. (Original post by jetskiwavedunkno)
for each curve find the gradient at the given point.

remember that the constant [ie: number without the x next it] differentiated gives 0.
13. (Original post by mr tim)
remember that the constant [ie: number without the x next it] differentiated gives 0.
for each curve find the gradient at the given point.

y= x^3-2 at x=1 gradient= 3

are they right
14. (Original post by jetskiwavedunkno)
for each curve find the gradient at the given point.

how do i do this
Take of the value, set the value you obtain equal to zero, then substitute in .

Spoiler:
Show
First one would be , so follow on from there.
15. (Original post by Mr Inquisitive)
Take of the value, set the value you obtain equal to , then substitute in .

Spoiler:
Show
First one would be , so follow on from there.
for each curve find the gradient at the given point.

y= x^3-2 at x=1 gradient= 3

are they right
16. (Original post by jetskiwavedunkno)
for each curve find the gradient at the given point.

y= x^3-2 at x=1 gradient= 3

are they right
yes they are all right.
17. (Original post by jetskiwavedunkno)
for each curve find the gradient at the given point.

y= x^3-2 at x=1 gradient= 3

are they right
Yes, they are.
18. (Original post by jetskiwavedunkno)
you are wrong

(Original post by jordan-s)
I'm afraid the gradient at 3 is also incorrect?

3x^2 when x = 3 gives 3 multiplied by 3^2, which is 3 x 9 = 27..

Ooops, too late! You've got it now anyway.
Sorry, i also calculated it as 27 but i don't know why did i wrote the gradient at x=3 is correct. Actually i was busy at another thread as well so i just got confused.

Btw, thanks for indicating my mistake.
19. (Original post by raheem94)
Sorry, i also calculated it as 27 but i don't know why did i wrote the gradient at x=3 is correct. Actually i was busy at another thread as well so i just got confused.

Btw, thanks for indicating my mistake.
20. (Original post by Mr Inquisitive)
Yes, they are.
Thank you sir

Updated: February 23, 2012
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