Question

Can anyone please help

Answer 1

$\text{Hi there! :)}$

$\large\boxed{f'(2) = -\frac{16}{27} }$

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

Answer 2

Answer:

1) $f'(x)=-\frac{24x}{(5+x^2)^2}$

2) $f'(2)=-\frac{16}{27}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{7-x^2}{5+x^2}$

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:

$\frac{d}{dx}[f(x)]=\frac{d}{dx}[\frac{7-x^2}{5+x^2}]$

Remember that the quotient rule is:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

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

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

$f'(x)=\frac{\frac{d}{dx}[7-x^2](5+x^2)-(7-x^2)\frac{d}{dx}[5+x^2]}{(5+x^2)^2}$

Differentiate:

$f'(x)=\frac{(-2x)(5+x^2)-(7-x^2)(2x)}{(5+x^2)^2}$

Simplify. Distribute in the numerator:

$f'(x)=\frac{(-10x-2x^3)-(14x-2x^3)}{(5+x^2)^2}$

Distribute:

$f'(x)=\frac{(-10x-2x^3)+(-14x+2x^3)}{(5+x^2)^2}$

The cubed terms cancel. This leaves:

$f'(x)=\frac{(-10x)+(-14x)}{(5+x^2)^2}$

Add. So, our derivative is:

$f'(x)=-\frac{24x}{(5+x^2)^2}$

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

$f'(2)=-\frac{24(2)}{(5+(2)^2)^2}$

Multiply and square:

$f'(2)=-\frac{48}{(5+4)^2}$

Add:

$f'(2)=-\frac{48}{(9)^2}$

Square:

$f'(2)=-\frac{48}{81}$

Reduce by 3:

$f'(2)=-\frac{16}{27}$

And we're done!