If f x -2 x -3- Then f - x -A- 1-2 x-3-2 -2 x-3C- -2 x-32D- 2-2 x-3

The correct option is A 1/√(2x+3)f(x) = √(2x+3) f'(x)=lim_h → 0f(x+h) f(x)/h f'(x)= lim_h → 0√(2(x+h)+3) √(2x+3)/h Multiply the numerator and denominator with ...

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Byju's Answer

Standard XI

Mathematics

Law of Reciprocal

If f x =√2 x ...

Question

If $f (x) = \sqrt{2 x + 3}$ , Then $f^{'} (x) =$ ___

\sqrt[3]{(2 x + 3)^{2}}

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\frac{1}{2 \sqrt{(2 x + 3)}}

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\frac{1}{\sqrt{2 x + 3}}

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\frac{2}{\sqrt{2 x + 3}}

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Solution

The correct option is C $\frac{1}{\sqrt{2 x + 3}}$
$f (x) = \sqrt{2 x + 3} f^{'} (x) = {lim}_{h \to 0} \frac{f (x + h) - f (x)}{h} f^{'} (x) = {lim}_{h \to 0} \frac{\sqrt{2 (x + h) + 3} - \sqrt{2 x + 3}}{h}$
Multiply the numerator and denominator with
$(\sqrt{2 x + 2 h + 3} + \sqrt{2 x + 3}) f^{'} (x) = {lim}_{h \to 0} \frac{(2 x + 2 h + 3 - 2 x - 3)}{h (\sqrt{2 x + 2 h + 3} + \sqrt{2 x + 3})} f^{'} (x) = {lim}_{h \to 0} \frac{2 h}{h} . {lim}_{h \to 0} (\frac{1}{\sqrt{2 x + 2 h + 3} + \sqrt{2 x + 3}}) f^{'} (x) = 2. \frac{1}{2 \sqrt{2 x + 3}} = \frac{1}{\sqrt{2 x + 3}}$

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If f(x)=√2x+3, Then f′(x)=___

If $f (x) = \sqrt{2 x + 3}$ , Then $f^{'} (x) =$ ___