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

Latest posts by Athena Estudy (see all)