(A) Let $u = \sqrt{x - 1}$. Since $1 < x < 26$,we have $0 < u < 5$.Then $x - 1 = u^2$,so $x = u^2 + 1$.The function becomes $f(x) = u + \sqrt{u^2 + 1

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(A) Let $u = \sqrt{x - 1}$. Since $1 Then $x - 1 = u^2$,so $x = u^2 + 1$.The function becomes $f(x) = u + \sqrt{u^2 + 1 + 24 - 10u} = u + \sqrt{u^2 - 10u + 25}$.This simplifies to $f(x) = u + \sqrt{(u - 5)^2} = u + |u - 5|$.Since $0 Thus,$f(x) = u + 5 - u = 5$.Since $f(x) = 5$ is a constant function for $1 < x < 26$,its derivative $f'(x) = 0$ for all $x$ in the interval $(1, 26)$.

Class 12 Mathematics Continuity and Differentiation Derivative at a point, Standard differentiation

Medium

Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}$ for $1 < x < 26$ be a real-valued function. Then $f'(x)$ for $1 < x < 26$ is

A
$0$
B
$\frac{1}{\sqrt{x - 1}}$
C
$2\sqrt{x - 1} - 5$
D
None of these

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Derivative at a point, Standard differentiation

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Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}$ for $1 < x < 26$ be a real-valued function. Then $f'(x)$ for $1 < x < 26$ is

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Assertion: For $x < 0$, $\frac{d^2}{d x^2}(\log |x|) = \frac{1}{|x|^2}$.
Reason: For $x < 0$, $|x| = -x$.