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Chain Rule of Differentiation

If fx = x 2+6...

Question

If $f (x) = \sqrt{x^{2} + 6 x + 9}$ , then f '(x) is equal to

A

1 f o r x > - 3

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B

- 1 f o r x < - 3

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C

1 f o r a l l x \in R

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D

0 f o r x = - 3

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Solution

The correct options are
A $1 f o r x > - 3$
B $- 1 f o r x < - 3$
$f (x) = \sqrt{x^{2} + 6 x + 9} = (x + 3 | Function is not differentiable at x = - 3 . f (x) = (\begin{matrix} - (x + 3), if x < - 3 (x + 3), if x > - 3 \end{matrix} \Rightarrow f' (x) = (\begin{matrix} - 1, if x < - 3 1, if x > - 3 \end{matrix}$

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Similar questions

f (x) = \sqrt{x^{2} + 6 x + 9}, then f' (x)

is equal to
(a)

1 for x < - 3

- 1 for x < - 3

1 for all x \in R

(d) none of these

f (x) = |x^{2} - 9 x + 20|

, then f' (x) is equal to
(a)

- 2 x + 9 for all x \in R

2 x - 9 if 4 < x < 5

- 2 x + 9, if 4 < x < 5

(d) none of these

f (x) = \sqrt{x^{2} + 6 x + 9}

f^{'} (x)

If f(x) = x, x≤1, and f(x) =x² + bx + c, x>1, and f'(x) exists finitely for all x ϵ R, then

f (x) = x \int - 1 | t | d t

x \geq 0, f (x)

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