Question

hii guys
please show this sum...​

Answer 1

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

Differentiating with respect to x,

$\frac{dy}{dx} = \frac{d}{dx}( log( \sqrt{2x + \sqrt{4 {x}^{2} + {a}^{2} } } ) )$

Let,

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

So,

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

Now,

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

Let

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

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

Now,

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

So,

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

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

Answer 2

Find:

$\dfrac{dy}{dx}\left(log \left\{\sqrt{2x+\sqrt{4x^{2}+a^{2}}}\right\}\right)$

Solution:

$\dfrac{dy}{dx}(log \{\sqrt{2x+\sqrt{4x^{2}+a^{2}}}\})\\\\= \dfrac{1}{ \sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times \dfrac{dy}{dx} (\sqrt{2x+\sqrt{4x^{2}+a^{2}}})\\\\= \dfrac{1}{ \sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times \dfrac{1}{2\sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times \dfrac{dy}{dx}(2x+\sqrt{4x^{2}+a^{2}}) \\\\= \dfrac{1}{ \sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times \dfrac{1}{2\sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times 2 + \dfrac{1}{2\sqrt{4x^{2}+a^{2}}} \times \dfrac{dy}{dx}(4x^{2}+a^{2})$

$\\\\= \dfrac{1}{ \sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times \dfrac{1}{2\sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times 2 + \dfrac{1}{2\sqrt{4x^{2}+a^{2}}} \times 4(2x^{(2-1)})+0\\\\= \dfrac{1}{ \sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times \dfrac{1}{2\sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times 2 + \dfrac{1}{2\sqrt{4x^{2}+a^{2}}} \times 4(2x)\\\\= \dfrac{1}{ \sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times \dfrac{1}{2\sqrt{2x+\sqrt{4x^{2}+a^{2}}}} \times 2 + \dfrac{1}{2\sqrt{4x^{2}+a^{2}}} \times 8x$

$\\\\= \dfrac{1}{2} \times \dfrac{1}{ 2x+\sqrt{4x^{2}+a^{2}}} \times 2 + \dfrac{8x}{2\sqrt{4x^{2}+a^{2}}} \\\\= \dfrac{1}{2} \times \dfrac{1}{ 2x+\sqrt{4x^{2}+a^{2}}} \times \dfrac{4\sqrt{4x^{2}+a^{2}} + 8x}{2\sqrt{4x^{2}+a^{2}}}\\\\= \dfrac{1}{2} \times \dfrac{1}{ 2x+\sqrt{4x^{2}+a^{2}}} \times \dfrac{4(\sqrt{4x^{2}+a^{2}} + 2x)}{2\sqrt{4x^{2}+a^{2}}}\\\\= \dfrac{1}{ 2x+\sqrt{4x^{2}+a^{2}}} \times \dfrac{\sqrt{4x^{2}+a^{2}} + 2x}{\sqrt{4x^{2}+a^{2}}}\\\\= \dfrac{1}{\sqrt{4x^{2}+a^{2}}}$

Formulas Used:

$1. \: \dfrac{dy}{dx} \: logx = \dfrac{1}{x}\\\\2. \: \dfrac{dy}{dx} \: \sqrt{x} = \dfrac{1}{2\sqrt{x}}\\\\3. \: \dfrac{dy}{dx} \: x^{n} = n.x^{(n-1)}\\\\4. \: \dfrac{dy}{dx} \: k = 0$