#### Answer

$$\frac{{f'\left( {\sqrt {g\left( x \right)} } \right)\left( {g'\left( x \right)} \right)}}{{2\sqrt {g\left( x \right)} }}$$

#### Work Step by Step

$$\eqalign{
& \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) \cr
& {\text{Let }}f\left( x \right){\text{ and }}g\left( x \right){\text{ differentiable functions with }}g\left( x \right) \geqslant 0 \cr
& {\text{use the chain rule}} \cr
& \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = \frac{d}{{dx}}\left[ {f\left( {\sqrt {g\left( x \right)} } \right)} \right]\frac{d}{{dx}}\left[ {\sqrt {g\left( x \right)} } \right] \cr
& {\text{rewrite }}\sqrt {g\left( x \right)} {\text{ as }}{\left( {g\left( x \right)} \right)^{1/2}} \cr
& \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = \frac{d}{{dx}}\left[ {f\left( {\sqrt {g\left( x \right)} } \right)} \right]\frac{d}{{dx}}\left[ {{{\left( {g\left( x \right)} \right)}^{1/2}}} \right] \cr
& {\text{use the chain rule again}} \cr
& \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = \frac{d}{{dx}}\left[ {f\left( {\sqrt {g\left( x \right)} } \right)} \right]\left[ {\frac{1}{2}{{\left( {g\left( x \right)} \right)}^{ - 1/2}}} \right]\frac{d}{{dx}}\left[ {g\left( x \right)} \right] \cr
& {\text{simplifying}} \cr
& \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = f'\left( {\sqrt {g\left( x \right)} } \right)\left[ {\frac{1}{2}{{\left( {g\left( x \right)} \right)}^{ - 1/2}}} \right]\left( {g'\left( x \right)} \right) \cr
& \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = f'\left( {\sqrt {g\left( x \right)} } \right)\left[ {\frac{1}{{2\sqrt {g\left( x \right)} }}} \right]\left( {g'\left( x \right)} \right) \cr
& \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = \frac{{f'\left( {\sqrt {g\left( x \right)} } \right)\left( {g'\left( x \right)} \right)}}{{2\sqrt {g\left( x \right)} }} \cr} $$