Question

Find f'(x) from the function

Answer 1

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

Explanation:

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

To find f'(x):

We know that
`(d)/(dx) √(x) =(1)/(2√(x))`

Now,

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

`f'(x)=\frac{1}{2\sqrt{25-x^(2)} } (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