If $f ( x ) = \sqrt { x + \sqrt { x } } , \text { find } f ^ { \prime } ( 1 )$
A)
$\frac { 3 \sqrt { 2 } } { 8 }$
B) $\frac { \sqrt { 2 } } { 4 }$
C) $\frac { 1 } { 4 }$
D) $\frac { 1 } { 8 }$
E) $\frac { \sqrt { 2 } } { 2 }$
F) $\frac { 1 } { 2 }$
G) $\sqrt { 2 }$
H)1

Question

If $f ( x ) = \sqrt { x + \sqrt { x } } , \text { find } f ^ { \prime } ( 1 )$ 
A)
$\frac { 3 \sqrt { 2 } } { 8 }$
B) $\frac { \sqrt { 2 } } { 4 }$
C) $\frac { 1 } { 4 }$
D) $\frac { 1 } { 8 }$
E) $\frac { \sqrt { 2 } } { 2 }$
F) $\frac { 1 } { 2 }$
G) $\sqrt { 2 }$
H)1

Quizplus · Accepted Answer

To find $f'(1)$, we first differentiate $f(x) = \sqrt{x + \sqrt{x}}$ with respect to $x$. Using the chain rule, $f'(x) = \frac{1}{2\sqrt{x + \sqrt{x}}} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right)$. Substituting $x = 1$, $f'(1) = \frac{1}{2\sqrt{1 + \sqrt{1}}} \cdot \left(1 + \frac{1}{2\sqrt{1}}\right) = \frac{1}{2\sqrt{2}} \cdot \frac{3}{2} = \frac{3\sqrt{2}}{8}$.

If f(x)=x+x, find f′(1)f ( x ) = \sqrt { x + \sqrt { x } } , \text { find } f ^ { \prime } ( 1 )f(x)=x+x​​, find f′(1)

If $f ( x ) = \sqrt { x + \sqrt { x } } , \text { find } f ^ { \prime } ( 1 )$