Question

Can anyone please help

Answer 1

1)
`f'(x)=-(24x)/((5+x^2)^2)`

2)
`f'(2)=-(16)/(27)`

Explanation:

So we have the function:

`f(x)=(7-x^2)/(5+x^2)`

And we want to find f'(x).

To do so, we can use the quotient rule.

So, let's take the derivative of both sides:

`(d)/(dx)[f(x)]=(d)/(dx)[(7-x^2)/(5+x^2)]`

Remember that the quotient rule is:

`(d)/(dx)[f/g]=(f'g-fg')/(g^2)`

In our equation, f is (7-x^2) and g is (5+x^2).

So, using the quotient rule, our derivative f'(x) is:

`f'(x)=((d)/(dx)[7-x^2](5+x^2)-(7-x^2)(d)/(dx)[5+x^2])/((5+x^2)^2)`

Differentiate:

`f'(x)=((-2x)(5+x^2)-(7-x^2)(2x))/((5+x^2)^2)`

Simplify. Distribute in the numerator:

`f'(x)=((-10x-2x^3)-(14x-2x^3))/((5+x^2)^2)`

Distribute:

`f'(x)=((-10x-2x^3)+(-14x+2x^3))/((5+x^2)^2)`

The cubed terms cancel. This leaves:

`f'(x)=((-10x)+(-14x))/((5+x^2)^2)`

Add. So, our derivative is:

`f'(x)=-(24x)/((5+x^2)^2)`

To find f'(2), simply substitute 2 into our derivative. So:

`f'(2)=-(24(2))/((5+(2)^2)^2)`

Multiply and square:

`f'(2)=-(48)/((5+4)^2)`

Add:

`f'(2)=-(48)/((9)^2)`

Square:

`f'(2)=-(48)/(81)`

Reduce by 3:

`f'(2)=-(16)/(27)`

And we're done!

Answer 2

`\text{Hi there! :)}`

`\large\boxed{f'(2) = -(16)/(27) }`

`f'(x) = ((5 + x^(2) )(-2x) - (7 - x^(2) )(2x))/((5+x^(2))^(2) ) \\\\f'(x) = (-10x-2x^(3)- 14x + 2x^(3) )/((5+x^(2))^(2) ) \\\\f'(x) = (-24x)/((5+x^(2))^(2) )\\\\ \text{Solve for the derivative at f'(2) using substitution:}\\\\f'(2) = (-24(2))/((5+2^(2))^(2) ) \\\\f'(2) = (-48)/(81) \\\\\text{Simplify:}\\\\f'(2) = -(16)/(27)`