# differentiation

• Feb 9th 2010, 12:54 AM
Awsom Guy
differentiation
Is this differentiation right.
question:
(2x-5)^3

6(2x-5)^2

If I do this how do I find the gradient. If I sub x=0 i don't get what I want.
• Feb 9th 2010, 01:04 AM
Quote:

Originally Posted by Awsom Guy
Is this differentiation right.
question:
(2x-5)^3

6(2x-5)^2

If I do this how do I find the gradient. If I sub x=0 i don't get what I want.

The differentiation is correct.
That is the curve gradient for any x.
If you sub in x=0, you get the slope of the tangent to the curve at x=0,
but not the slope of the tangent that passes through the origin if that's what you were looking for.

What were you trying to calculate ?
• Feb 9th 2010, 01:05 AM
Prove It
Quote:

Originally Posted by Awsom Guy
Is this differentiation right.
question:
(2x-5)^3

6(2x-5)^2

If I do this how do I find the gradient. If I sub x=0 i don't get what I want.

I assume you have used the chain rule to arrive at this answer.

When you sub $\displaystyle x = 0$ into the derivative, you should find that the gradient at that point is $\displaystyle 150$.
• Feb 9th 2010, 01:06 AM
Awsom Guy
My question says: Find the point where the curve y=(2x-5)^3 cuts the x-axis and find the slope of the tangent at this point.
• Feb 9th 2010, 01:09 AM
Prove It
Quote:

Originally Posted by Awsom Guy
My question says: Find the point where the curve y=(2x-5)^3 cuts the x-axis and find the slope of the tangent at this point.

Well then do as you are instructed.

The find where the curve cuts the $\displaystyle x$ axis, let $\displaystyle y = 0$.

So $\displaystyle 0 = (2x - 5)^3$

$\displaystyle 2x - 5 = 0$

$\displaystyle x = \frac{5}{2}$.

So you are not evaluating the derivative at $\displaystyle x = 0$. You are evaluating the derivative at $\displaystyle x = \frac{5}{2}$.
• Feb 9th 2010, 01:12 AM
Awsom Guy
do i then place it in the y=mx+b formula to calculate the slope at the tagent at the point.
• Feb 9th 2010, 01:14 AM
Awsom Guy
just so everyone knows the answer is m=0
• Feb 9th 2010, 01:22 AM
no need,
the derivative is the slope or gradient of the tangent.
That's zero when y=0 also in this case.
• Feb 9th 2010, 01:24 AM
Awsom Guy
But the derivative isn't 0 it is 6(2x-5)^2
• Feb 9th 2010, 01:27 AM
It is when $\displaystyle x=\frac{5}{2}$ or y=0 !