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Mathematics 8 Online
OpenStudy (anonymous):

calculus problem

zepdrix (zepdrix):

\[\Large\rm v(x)=f(f(x))\]By the definition of the chain rule:\[\Large\rm v'(x)=f'(f(x))\cdot f'(x)\]

zepdrix (zepdrix):

\[\Large\rm v'(1)=f'(\color{orangered}{f(1)})\cdot f'(1)\]Deal with the orange part first.\[\Large\rm f(1)=?\]

OpenStudy (anonymous):

1?

OpenStudy (anonymous):

and derivative of 1 is 0 and for the other part 3 is also 0 .. so f(1) = 1 f(3) = 2 ... ( from the graph) ?

OpenStudy (ranga):

What is f(1) from the graph?

OpenStudy (anonymous):

2

OpenStudy (anonymous):

@ranga

OpenStudy (ranga):

Yes. f(1) = 2. Put that in: f'(f(1)) as shown by zepdrix above.

OpenStudy (anonymous):

isnt the derivative of 2 zero?

OpenStudy (ranga):

You need to find f'(f(1)). That is, f'(2). f'(2) is the slope of the graph f(x) at x = 2. What is the slope of f(x) at x = 2?

OpenStudy (anonymous):

the slope doesnt exist cause its a corner... right

OpenStudy (ranga):

Exactly!

OpenStudy (anonymous):

okay so for v'(3) = would be f'(2) aswell so DNE aswell?

OpenStudy (ranga):

Yes!

OpenStudy (anonymous):

ty!

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