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Differentiate

\(g\left( x \right) = \left( {x + {\bf{2}}\sqrt x } \right){e^x}\)

Short Answer

Expert verified

The derivative of g is \({e^x}\left( {x + 2\sqrt x + \frac{1}{{\sqrt x }} + 1} \right)\).

Step by step solution

01

Step 1:Find the derivative of f using the product rule

The equation forproduct rule is \(\left( {fg} \right)' = fg' + gf'\).

Apply Quotient rule for the function \(g\left( x \right) = \left( {x + 2\sqrt x } \right){e^x}\).

\(\begin{aligned}g'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\left( {x + 2\sqrt x } \right){e^x}} \right)\\ &= \left( {x + 2\sqrt x } \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^x}} \right) + \left( {{e^x}} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x + 2\sqrt x } \right)\end{aligned}\)

02

Differentiate the equation in step 1

The derivative of g can be obtained as,

\(\begin{aligned}g'\left( x \right) &= \left( {x + 2\sqrt x } \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^x}} \right) + \left( {{e^x}} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x + 2\sqrt x } \right)\\ &= {e^x}\left( {x + 2\sqrt x } \right) + {e^x}\left( {1 + \frac{1}{{\sqrt x }}} \right)\\ &= {e^x}\left( {x + 2\sqrt x + \frac{1}{{\sqrt x }} + 1} \right)\end{aligned}\)

Thus, the derivative of gis \({e^x}\left( {x + 2\sqrt x + \frac{1}{{\sqrt x }} + 1} \right)\).

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