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need help on dy/dx

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Original post by Mr Inquisitive
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.


is it x=3 graient =27
x=5 gradient =75

x=-3 gradient =27??
Reply 21
Original post by jetskiwavedunkno
is it x=3 graient =27
x=5 gradient =75

x=-3 gradient =27??


Those are all right.
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
Original post by Gemini92
Those are all right.


awww thank you! xx
Original post by jetskiwavedunkno
is it x=3 graient =27
x=5 gradient =75

x=-3 gradient =27??


Yes, they're correct.
Reply 25
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.
(edited 12 years ago)
Original post by Mr Inquisitive
Yes, they're correct.


for each curve find the gradient at the given point.

y=4x^2+2 at x=1 gradient=

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

y= 4x^3 at x=1 gradient=

how do i do this
Reply 27
Original post by jetskiwavedunkno
for each curve find the gradient at the given point.

y=4x^2+2 at x=1 gradient=

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

y= 4x^3 at x=1 gradient=

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)
(edited 12 years ago)
Reply 28
Original post by jetskiwavedunkno
for each curve find the gradient at the given point.

y=4x^2+2 at x=1 gradient=

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

y= 4x^3 at x=1 gradient=

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
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)




y=4x^2+2 at x=1 gradient=10??
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.

y=4x^2+2 at x=1 gradient=10??
Reply 31
Original post by jetskiwavedunkno
for each curve find the gradient at the given point.

y=4x^2+2 at x=1 gradient=10??


:no: remember that the constant [ie: number without the x next it] differentiated gives 0.
Original post by mr tim
:no: 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=4x^2+2 at x=1 gradient= 8

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

y= 4x^3 at x=1 gradient=12

are they right
(edited 12 years ago)
Original post by jetskiwavedunkno
for each curve find the gradient at the given point.

y=4x^2+2 at x=1 gradient=

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

y= 4x^3 at x=1 gradient=

how do i do this


Take dydx\frac{dy}{dx} of the yy value, set the value you obtain equal to zero, then substitute in x=1x=1.

Spoiler

(edited 12 years ago)
Original post by Mr Inquisitive
Take dydx\frac{dy}{dx} of the yy value, set the value you obtain equal to 00, then substitute in x=1x=1.

Spoiler



for each curve find the gradient at the given point.

y=4x^2+2 at x=1 gradient= 8

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

y= 4x^3 at x=1 gradient=12

are they right
Reply 35
Original post by jetskiwavedunkno
for each curve find the gradient at the given point.

y=4x^2+2 at x=1 gradient= 8

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

y= 4x^3 at x=1 gradient=12

are they right


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

y=4x^2+2 at x=1 gradient= 8

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

y= 4x^3 at x=1 gradient=12

are they right


Yes, they are.
Reply 37
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.
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.


dw about it raheem
Original post by Mr Inquisitive
Yes, they are.


Thank you sir

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