Use the rules for derivatives to find the derivative of each function defined as

Use the rules for derivatives to find the derivative of each function defined as follows.
$$\displaystyle{y}={x}\sqrt{{{x}}}+{\frac{{{3}}}{{{x}^{{{2}}}\sqrt{{{x}}}}}}$$

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Step 1
To find the derivative of: $$\displaystyle{y}={x}\sqrt{{{x}}}+{\frac{{{3}}}{{{x}^{{{2}}}\sqrt{{{x}}}}}}$$
solution:
$$\displaystyle{y}={x}\sqrt{{{x}}}+{\frac{{{3}}}{{{x}^{{{2}}}\sqrt{{{x}}}}}}$$
On simplifying further, we get:
$$\displaystyle{y}={x}\sqrt{{{x}}}+{\frac{{{3}}}{{{x}^{{{2}}}\sqrt{{{x}}}}}}$$
$$\displaystyle{y}={x}^{{{\frac{{{3}}}{{{2}}}}}}+{\frac{{{3}}}{{{x}^{{{\frac{{{5}}}{{{2}}}}}}}}}$$ (using, $$\displaystyle{a}^{{{b}}}{a}^{{{c}}}={a}^{{{b}+{c}}}$$)
$$\displaystyle{y}={x}^{{{\frac{{{3}}}{{{2}}}}}}+{3}{x}^{{{\frac{{-{5}}}{{{2}}}}}}$$ (using, $$\displaystyle{\frac{{{1}}}{{{a}^{{{n}}}}}}={a}^{{-{n}}}$$)
differentiating both sides w.r.t x we get:
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({y}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{\frac{{{3}}}{{{2}}}}}}+{3}{x}^{{{\frac{{-{5}}}{{{2}}}}}}\right)}$$
$$\displaystyle\Rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\left({\frac{{{3}}}{{{2}}}}{x}^{{{\frac{{{3}}}{{{2}}}}-{1}}}\right)}+{3}{\left(-{\frac{{{5}}}{{{2}}}}{x}^{{{\frac{{-{5}}}{{{2}}}}-{1}}}\right)}$$ (using, $$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{n}}}\right)}={n}{x}^{{{n}-{1}}}$$)
$$\displaystyle\Rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{3}}}{{{2}}}}\sqrt{{{x}}}-{\frac{{{15}}}{{{2}}}}{x}^{{-{\frac{{{7}}}{{{2}}}}}}$$
$$\displaystyle\Rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{3}}}{{{2}}}}\sqrt{{{x}}}-{\frac{{{15}}}{{{2}{x}^{{{3}}}\sqrt{{{x}}}}}}$$
Step 2
Result:
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{3}}}{{{2}}}}\sqrt{{{x}}}-{\frac{{{15}}}{{{2}{x}^{{{3}}}\sqrt{{{x}}}}}}$$
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