Why would you use the definition of a derivative (lim h approaches ) f(a+h) - f(a)/h over the equivalent lim x approaches a f(x) - f(a)/ x-a? When finding the instantaneous rate of change at a point or on a interval.

Question

schrodingers_cat · Answer

Because you are trying to find the instantaneous rate of change at x. Both equations you wrote represents slopes just written in different forms. Just let f(b) = some value of x plus some value of h.

Does that make sense?

anonymous · Answer

I would work one using the bottom one and get it wrong, and then check my solution manual to find that they used the other form. Are there situations where one is more advantageous than the other?

schrodingers_cat · Answer

No they are used for different situations when take a derivative you let h go to zero. f(x) - f(a)/ x-a would just get the average rate of change between two points. Where as 

$$\lim_{h \rightarrow o} f(x+h) -f(x)/(h)$$

Will give you the slope of the line tangent to the function at any given point.

Does that make sense?

schrodingers_cat · Answer

They are both slopes but one gives the average rate of change and the derivative gives the instantaneous rate of change.

anonymous · Answer

So the average rate of change is used for an interval and the instantaneous rate of change for a point?

anonymous · Answer

So the definition of a derivative would be applicable to finding the slope of a tangent , but the other form would not? Or is the instantaneous the slope of the tangent at that point and therefore the values would be the same?

schrodingers_cat · Answer

Your first statement is correct. The values will not be the same. Essentially when you are taking a derivative you are doing this

|dw:1385519710593:dw|

You are essentially going through all values from h to zero until you get to zero the left most line would represent the slope of the tangent at a given point a.