stumpted! differentiation?

Question

how to you differentiate this y= (X)^3 / 3?
Please show working!
Thanks

Answer 1

y = x^3 / 3

When you differentiate x to some power, you simply bring the power to the front and subtract one from the power.

For example, if you had x^n, you would make this nx^(n-1).

In your case you have:
y = x^3 / 3
y' = 3 x^2 / 3

Then the threes cancel out and you have:
y' = x^2

Answer 2

please show working? once you understand differentiation you can solve this by inspection (in fact, there are many more difficult problems that can be solved by inspection. "by inspection" was the buzz word when i took a differential equations course).
if you want to show all work, start with the definition of the derivative,
you know, the limit as delta x approaches zero of [f(x + delta x) - f(x)]/ delta x and work your way through it.

but math is all about reducing the problem to a simpler case which you already know how to solve.

to differentiate y = x^3 / 3
you need to know two things:
1.
the derivative of a constant times a function is the constant times the derivative of the function
this applies here because we have 1/3 times x^3
2.
the derivative of x raised to a power n is n times x raised to the power n-1

so, using operator notation D() for differentiation with respect to x,
D(y) = D(x^3 / 3) = (1/3) * D(x^3) = (1/3) * 3x^(3-1) = x^2

Answer 3

y= (x^3)/3

is the same as:

y= 1/3(x^3)

when differentiated is:

dy/dx= (3 (the power) multiplied by 1/3 (the coefficient of x)) equals 1 so one is the resultant coefficient of x

Then minus 1 from the previous coefficent (3) to come up with an answer of dy/dx= x^2

Answer 4

Differentiation in both side:
dy = d( X^3 /3)
=1/3. d(X^3) ;as1/3 is constant,it will be places outside of the diff.
=1/3 (3 X^(3-1)) ;rule: d(Z^b)= b.Z^(b-1)
= X^2
so, answer is : X^2

Answer 5

when u differentiate both sides:
dy/dx= d/dx (x^3/3)

= 1/3 (3x^2) [since 1/3 is a constant u dont differentiate it]

[cancelling the two 3s]

= x^2
-----------------

Answer 6

y= x^3 / 3
dy/dx i.e. defferentiation of y w.r.t x
it will be 1/3 * 3x^2 (since 1/3 is constant so it can be taken out from defferentiation process)
now cancelling the threes.
final answer becomes
dy/dx = x^2

Answer 7

generally: if y=x^n then y'=(n)*x^(n-1) therefore: if y=(x)^3/3
then y'=3*x^(3-1)/3
=x^2 (3 and 3 cancel)