Question

Find f'(x) from the function

Answer 1

Answer:

f'(x)=$\frac{-x}{\sqrt{25-x^{2}} }$ for (-5)<x<5

Step-by-step explanation:

The given function is f(x)=$\sqrt{25-x^{2} }$ for $-5\leq x\leq 5$

To find f'(x):

We know that $\frac{d}{dx} \sqrt{x} =\frac{1}{2\sqrt{x}}$

Now,

f'(x)=$\frac{d}{dx}\sqrt{25-x^{2} }$

$f'(x)=\frac{1}{2\sqrt{25-x^{2}} } \frac{d}{dx} (25-x^{2})\\f'(x)=\frac{}{2\sqrt{25-x^{2}} } (0-2x)\\f'(x)=\frac{-2x}{2\sqrt{25-x^{2}} }\\f'(x)=\frac{-x}{\sqrt{25-x^{2}} }$

Such that for x=5 or x=(-5), f'(x) is indefinite

Thus,

f'(x)=$\frac{-x}{\sqrt{25-x^{2}} }$ for (-5)<x<5