1. ## 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.

2. 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 ?

3. 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 $x = 0$ into the derivative, you should find that the gradient at that point is $150$.

4. 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.

5. 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 $x$ axis, let $y = 0$.

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

$2x - 5 = 0$

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

So you are not evaluating the derivative at $x = 0$. You are evaluating the derivative at $x = \frac{5}{2}$.

6. do i then place it in the y=mx+b formula to calculate the slope at the tagent at the point.

7. just so everyone knows the answer is m=0

8. no need,
the derivative is the slope or gradient of the tangent.
That's zero when y=0 also in this case.

9. But the derivative isn't 0 it is 6(2x-5)^2

10. It is when $x=\frac{5}{2}$ or y=0 !

11. Thank you so much for helping me.

12. here's the graph,
the tangent is the x axis itself at x=2.5.

13. thanks