I got zero but apparently that isnt the answer? can some one explain to me why zero isnt the correct answer and how to properly solve this? I’m confused :S
If f(x)=x^n then f ‘(x)=nx^(n-1), which you should know.
Sub n=1 so f(x)=x^1=x then f ‘(x)=1x^0=1 since x^0=1
So we have f ‘(x)=1 and this is true for all x and so f ‘(5)=1.
If you sketch the graph of y=x the value of f ‘(5) is the slope of
the graph at x=5 and this is clearly 1.
f'(x)=1
f'(5)=1
f (x) = x
f ` (x) = 1
f ` (5) = 1
y = f (x) = x is the equation of a straight line that has gradient m = 1 and passes thro` the origin
f(x)=x —> f'(x)=1 , hence f'(5)=1
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