Question

Can someone please help with this math problem

Answer 1

Given:

The function is:

`y=(f(x))^2`

To find:

The derivative of the given function at x=1, i.e., y'(1).

Solution:

Chain rule of differentiation:

`[f(g(x))]'=f'(g(x))g'(x)`

We have,

`y=(f(x))^2`

Differentiate with respect to x.

`y'(x)=2(f(x))(d)/(dx)f(x)`

`y'(x)=2f(x)f'(x)`

Putting
`x=1`, we get

`y'(1)=2f(1)f'(1)`

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

`y'(1)=2(3)(3)`

`y'(1)=18`

Therefore, the value of y'(1) is 18.