Simple differentition

[wellY--3]

hi,

I understand how to use the limiting chord process to find the gradient of a curve, but how do you do it when they ask for a certain point on the curve. for example

find the gradient using the limiting chord process on the curve y=x^2 +2 at the point (1,3)

If you sub the points into the equation it's no longer y= which you need to be able to use the formula they're wanting

 

[HallsofIvy]

Nothing to do with differential equations! I am moving this to "Calculus and Analysis".

"find the gradient using the limiting chord process on the curve y=x^2 +2 at the point (1,3)."

The gradient is given by
$$\lim_{h\rightarrow 0}\frac{f(1+h)-f(1)}{h}$$

With f(x)= x²+ 2, what is f(1)? What is f(1+h)?

 

[tacman]

you to use


Hi there,

When finding the gradient using the limiting chord process at a specific point on the curve, you will still use the same formula. In this case, the formula would be: (f(x+h) - f(x)) / h, where h represents the change in x. To find the gradient at the point (1,3), you would substitute x=1 and y=3 into the formula. This will give you the gradient at that specific point on the curve. Remember that the value of h is very small and approaches 0, which is why we use the limiting process to find the gradient. I hope this helps!