Verified answer

You must have made a mistake using the definition of the

gradient { we call it the derivative here in United States

in most classes but it is same idea}

Since y = 2x³

The derivative {gradient} is calculated by

using the DIFFERENCE QUOTIENT

then evaluating the limit as h→0

................f(x+h) - f(x)

y' =f'(x) = -------------------- as h approaches zero

......................h

the difference quotient is

2(x+ h)³ - 2x³

--------------------

......h

you did figure out that

(x+ h)³ = x³ + 3x²h + 3xh² + h³

I think so

2(x+ h)³ - 2x³

--------------------

......h

is equal to

2(x³ + 3x²h + 3xh² + h³) - 2x³

--------------------------------------------------

................h

I THINK YOU MAY HAVE ERRORED HERE

IN NOT SHOWING THAT 2 must be multiplied by

the quantity in parentheses

that gives you

2x³ + 6x²h + 6xh² + 2h³ - 2x³

--------------------------------------------------

................h

The 2x³ 's all cancel so you get

6x²h + 6xh² + 2h³

-------------------------------

.............h

now factor h out of the numerator

then take the limit of this expression as h→0

From the definition, you should have 2(x+h)^3 - 2x^3 all over h. It looks like you didn't distribute the 2 to everything.

Then once you distribute the cube, this becomes 2x^3 + 6hx^2 + 6xh^2 + h^3 - 2x^3, all over h.

The 2x^3 and -2x^ cancel, leaving you with 6hx^2+ 6xh^2 +h^3, all over h. Then you can factor out and cancel an h, giving you 6x^2 + 6xh + h^2. Take the limit as h goes to 0, and you're left with 6x^2.

Dy/Dx= (2(x+h)^3 -2x^3)/h

=( 2(x^3+3x^2h+3xh^2+h^3) -2x^3) /h

= (6x^2h+6xh^2 +2h^3)/h

=h((6x^2+6xh +2h^2)) /h

=((6x^2+6xh +2h^2))

Lim Dy/Dx h goes to 0 = 6x^2

There should be a "x^3" term also, which cancels the "-x^3". Then you can cancel out a common factor of h and the limit is then trivial.

( 2(x+h)³ - 2x³ ) / h

= (2h³ + 6h²x + 6hx²) / h

= 2h² + 6xh + 6x²

As h->0, the above becomes: 6x²