Given:
The function is:
[tex]y=(f(x))^2[/tex]
To find:
The derivative of the given function at x=1, i.e., y'(1).
Solution:
Chain rule of differentiation:
[tex][f(g(x))]'=f'(g(x))g'(x)[/tex]
We have,
[tex]y=(f(x))^2[/tex]
Differentiate with respect to x.
[tex]y'(x)=2(f(x))\dfrac{d}{dx}f(x)[/tex]
[tex]y'(x)=2f(x)f'(x)[/tex]
Putting [tex]x=1[/tex], we get
[tex]y'(1)=2f(1)f'(1)[/tex]
From the given table of values it is clear that f(1)=3 and f'(1)=3. Putting these values in the above equation, we get
[tex]y'(1)=2(3)(3)[/tex]
[tex]y'(1)=18[/tex]
Therefore, the value of y'(1) is 18.
