Question

y=log (sqrt x+sqrt x-a)then dy/dx is​

Answer 1

$Given \: y = log (\sqrt{x} + \sqrt{x-a})$

/* Differentiating with respect to 'x' on bothsides of the equation, we get */

$\frac{d}{dx}(y) = \frac{d}{dx} [ log (\sqrt{x} + \sqrt{x-a}]$

$= \frac{1}{ (\sqrt{x} + \sqrt{x-a}) } \times \frac{d}{dx} (\sqrt{x} + \sqrt{x-a})$

$\boxed{\pink{ \because \frac{d}{dx} \{ log [f(x)]\} = \frac{1}{f(x)} \frac{d}{dx} [f(x)] }}$

$= \frac{1}{ (\sqrt{x} + \sqrt{x-a}) } \times [ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{x-a}} ]$

$= \cancel {\frac{1}{ (\sqrt{x} + \sqrt{x-a}) }} \times \frac{1}{2} [ \frac{\cancel {\sqrt{x-a}+ \sqrt{x}}}{\sqrt{x}\times (\sqrt{x-a})}]$

$= \frac{1}{2\sqrt{x}(\sqrt{x-a})}$

$\green {= \frac{1}{2\sqrt{x^{2} - ax}} }$

•••♪

Answer 2

Giveny=log(

x

+

x−a

)

/* Differentiating with respect to 'x' on bothsides of the equation, we get */

\frac{d}{dx}(y) = \frac{d}{dx} [ log (\sqrt{x} + \sqrt{x-a}]

dx

d

(y)=

dx

d

[log(

x

+

x−a

]

= \frac{1}{ (\sqrt{x} + \sqrt{x-a}) } \times \frac{d}{dx} (\sqrt{x} + \sqrt{x-a})=

(

x

+

x−a

)

1

×

dx

d

(

x

+

x−a

)

\boxed{\pink{ \because \frac{d}{dx} \{ log [f(x)]\} = \frac{1}{f(x)} \frac{d}{dx} [f(x)] }}

∵

dx

d

{log[f(x)]}=

f(x)

1

dx

d

[f(x)]

= \frac{1}{ (\sqrt{x} + \sqrt{x-a}) } \times [ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{x-a}} ]=

(

x

+

x−a

)

1

×[

2

x

1

+

2

x−a

1

]

= \cancel {\frac{1}{ (\sqrt{x} + \sqrt{x-a}) }} \times \frac{1}{2} [ \frac{\cancel {\sqrt{x-a}+ \sqrt{x}}}{\sqrt{x}\times (\sqrt{x-a})}]=

(

x

+

x−a

)

1

×

2

1

[

x

×(

x−a

)

x−a

+

x

]

= \frac{1}{2\sqrt{x}(\sqrt{x-a})}=

2

x

(

x−a

)

1

\green {= \frac{1}{2\sqrt{x^{2} - ax}} }=

2

x

2

−ax

1