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:

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{3}}}{{{2}}}}\sqrt{{{x}}}-{\frac{{{15}}}{{{2}{x}^{{{3}}}\sqrt{{{x}}}}}}\)

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:

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{3}}}{{{2}}}}\sqrt{{{x}}}-{\frac{{{15}}}{{{2}{x}^{{{3}}}\sqrt{{{x}}}}}}\)