Answer:
f'(x)=[tex]\frac{-x}{\sqrt{25-x^{2}} }[/tex] for (-5)<x<5
Step-by-step explanation:
The given function is f(x)=[tex]\sqrt{25-x^{2} }[/tex] for [tex]-5\leq x\leq 5[/tex]
To find f'(x):
We know that [tex]\frac{d}{dx} \sqrt{x} =\frac{1}{2\sqrt{x}}[/tex]
Now,
f'(x)=[tex]\frac{d}{dx}\sqrt{25-x^{2} }[/tex]
[tex]f'(x)=\frac{1}{2\sqrt{25-x^{2}} } \frac{d}{dx} (25-x^{2})\\f'(x)=\frac{}{2\sqrt{25-x^{2}} } (0-2x)\\f'(x)=\frac{-2x}{2\sqrt{25-x^{2}} }\\f'(x)=\frac{-x}{\sqrt{25-x^{2}} }[/tex]
Such that for x=5 or x=(-5), f'(x) is indefinite
Thus,
f'(x)=[tex]\frac{-x}{\sqrt{25-x^{2}} }[/tex] for (-5)<x<5
