Question

Q1.hii guys
please show this sum...


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Answer 1

Step-by-step explanation:

$y = ( log( \sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} } } ) </p><p></p><p>$

Differentiating with respect to x,

Let,

$b = \sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} } }$

So,

$\begin{gathered} \frac{dy}{dx} = \frac{d}{dx} (logb) \\ \\ \frac{ dy}{dx} = \frac{1}{b} \times \frac{ db}{dx} \end{gathered}$

Now,

$\frac{db}{dx} = \frac{d}{dx}( \sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} } }) </p><p>$

Let

$c = 2x + \sqrt{4 {x}^{2} + {a}^{2} } </p><p>$

$\frac{db}{dx} = \frac{d}{dx} \sqrt{c} = \frac{1}{2 \sqrt{c} } \times \frac{dc}{dx}$

Now,

$\begin{gathered} \frac{dc}{dx} = \frac{d}{dx} (2x + \sqrt{4 {x}^{2} + {a}^{2} }) \\ \\ \frac{dc}{dx} = 2 + \frac{8x}{2 \sqrt{ 4{x}^{2} + {a}^{2} } } \end{gathered}$

So,

$\begin{gathered} \frac{ dy}{dx} = \frac{1}{b} \times \frac{ db}{dx} \\ \\ \frac{ dy}{dx} = \frac{1}{\sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} } } } \times \frac{1}{2 \sqrt{c} } \times \frac{dc}{dx} \\ \\ \frac{ dy}{dx} = \frac{1}{\sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} } } } \times \frac{1}{2 \sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} }} } \times \frac{dc}{dx} \\ \\\frac{ dy}{dx} = \frac{1}{\sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} } } } \times \frac{1}{2 \sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} }} } \times ( 2 + \frac{8x}{2 \sqrt{ 4{x}^{2} + {a}^{2} } } )\end{gathered}$

$\begin{gathered} \\\frac{ dy}{dx} = \frac{1}{2(2x + \sqrt{4 {x}^{2} + {a}^{2} )} } \times ( 2 + \frac{8x}{2 \sqrt{ 4{x}^{2} + {a}^{2} } } ) \\ \\\frac{ dy}{dx} = \frac{1}{2(2x + \sqrt{4 {x}^{2} + {a}^{2} )} } \times ( \frac{2 (2 \sqrt{ 4{x}^{2} + {a}^{2} }) + 8x}{2 \sqrt{ 4{x}^{2} + {a}^{2} } } ) \\ \\ \\\frac{ dy}{dx} = \frac{1}{2(2x + \sqrt{4 {x}^{2} + {a}^{2} )} } \times ( \frac{4(\sqrt{ 4{x}^{2} + {a}^{2} }) + 2x}{2 \sqrt{ 4{x}^{2} + {a}^{2} } } ) \\ \\ \frac{ dy}{dx} = \frac{1}{2(2x + \sqrt{4 {x}^{2} + {a}^{2} )} } \times ( \frac{4(\sqrt{ 4{x}^{2} + {a}^{2} }+ 2x)}{2 \sqrt{ 4{x}^{2} + {a}^{2} } } ) \\ \\ \\ \frac{dy}{dx} = \frac{1}{ \sqrt{4 {x}^{2} + {a}^{2} } } \end{gathered}$