differentiation

Is this differentiation right.
question:
(2x-5)^3

My Answer:
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.
Please help.

Quote:

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Originally Posted by Awsom Guy

Is this differentiation right.
question:
(2x-5)^3

My Answer:
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.
Please help.

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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 ?

Quote:

---

Originally Posted by Awsom Guy

Is this differentiation right.
question:
(2x-5)^3

My Answer:
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.
Please help.

---

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

If so, then yes, your answer is correct.

When you sub into the derivative, you should find that the gradient at that point is .

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.

Quote:

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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.

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Well then do as you are instructed.

The find where the curve cuts the axis, let .

So



.


So you are not evaluating the derivative at . You are evaluating the derivative at .

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

just so everyone knows the answer is m=0

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

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

It is when or y=0 !

Thank you so much for helping me.

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

thanks