# Thread: Easy Differentiation -- Double Checking

1. ## Easy Differentiation -- Double Checking

Find the co-ordinates of the points on the curves given by the following equation at which the gradient has the given value: y = x2 - 3x + 1, gradient = 1.

To solve this, I found the derivate of the function: dy/dx = 2x - 3 and equated it to 1 and solved for x, giving me x = 2. Is this correct? How do I present my answer in terms of coordinates? What if there are two answers for x?

Also, I'm asked in a different question to differentiate f(x) = 5 by first principles. I know that the answer is zero, but how do I show working out to prove it?

2. ## Re: Easy Differentiation -- Double Checking

Hey Fratricide.

If there were multiple answers for the tangent you would have at least a two or more degree polynomial so this guarantees that you have one solution.

If you present the points in (x,y) form then you should be OK with regards to getting full marks.

3. ## Re: Easy Differentiation -- Double Checking

What about differentiating f(x) = 5 by first principles?

4. ## Re: Easy Differentiation -- Double Checking

You will always get zero if you differentiate a constant.

Is this part of an example or question? If it is can you please show us what the question/example is?

5. ## Re: Easy Differentiation -- Double Checking

The question asks me to differentiate f(x) = 5 by first principles, showing all workings out. I know the answer will always be zero, but how am I supposed to show workings out? Do I just put a big fat zero?

Edit: I let f(x) = 5 = 5x0, which leaves me with f'(x) = lim(h->0) 0/h, i.e., 0/0, i.e., undefined?

6. ## Re: Easy Differentiation -- Double Checking

Use the definition f'(x) = lim h -> 0 [f(x+h) - f(x)]/h where f(x) = 5 for all x.

7. ## Re: Easy Differentiation -- Double Checking

Yes, but the answer ends up being 0/0 (as per my previous post). Never mind though, the problem was sorted.