Question

Differentiate w.r.t x
$\sf \: f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10 \sqrt{x - 1} }$
a) 0
b) 1/√x - 1
c) 2√(x - 1) - 5
d) None ​

Answer 1

Given:

$f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10 \sqrt{x - 1} }$

To find:

Derivative of f(x) w.r.t x

Calculation:

$f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10 \sqrt{x - 1} }$

$= > f(x) = \sqrt{x - 1} + \sqrt{(x - 1) - 10 \sqrt{x - 1} + 25}$

$= > f(x) = \sqrt{x - 1} + \sqrt{ {( \sqrt{x - 1)} }^{2} - 2 \times \sqrt{x - 1} \times 5 + {(5)}^{2} }$

$= > f(x) = \sqrt{x - 1} + \sqrt{ {( \sqrt{x - 1} - 5) }^{2} }$

$= > f(x) = \sqrt{x - 1} + \sqrt{ x - 1} - 5$

$= > f(x) = 2\sqrt{x - 1} - 5$

Now , differentiating w.r.t x

$= > \dfrac{d \{ f(x) \}}{dx} =2 \dfrac{d( \sqrt{x - 1} )}{dx} - \dfrac{d(5)}{dx}$

$= > \dfrac{d \{ f(x) \}}{dx} =2 ( \dfrac{1}{2 \sqrt{x - 1} } ) - 0$

$= > \dfrac{ d \{f(x) \}}{dx} =\dfrac{1}{\sqrt{x - 1} }$

So, final answer is:

$\boxed{ \bf{ \red{ \dfrac{ d \{f(x) \}}{dx} =\dfrac{1}{\sqrt{x - 1} } }}}$

Answer 2

Answer:

b

Step-by-step explanation:

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