Proceeding as in example (1) above, we get
(6) Differentiate y = x 3 1 + x 2 {\displaystyle y={\sqrt {\dfrac {x^{3}}{1+x^{2}}}}} .
We may write this
y = x 3 2 ( 1 + x 2 ) − 1 2 {\displaystyle y=x^{\tfrac {3}{2}}(1+x^{2})^{-{\tfrac {1}{2}}}} ;
d y d x = 3 2 x 1 2 ( 1 + x 2 ) − 1 2 + x 3 2 × d [ ( 1 + x 2 ) − 1 2 ] d x {\displaystyle {\frac {dy}{dx}}={\tfrac {3}{2}}x^{\tfrac {1}{2}}(1+x^{2})^{-{\tfrac {1}{2}}}+x^{\tfrac {3}{2}}\times {\frac {d\left[(1+x^{2})^{-{\tfrac {1}{2}}}\right]}{dx}}} .
Differentiating ( 1 + x 2 ) − 1 2 {\displaystyle (1+x^{2})^{-{\frac {1}{2}}}} , as shown in example (2) above, we get
d [ ( 1 + x 2 ) − 1 2 ] d x = − x ( 1 + x 2 ) 3 {\displaystyle {\frac {d\left[(1+x^{2})^{-{\tfrac {1}{2}}}\right]}{dx}}=-{\frac {x}{\sqrt {(1+x^{2})^{3}}}}} ;
so that
d y d x = 3 x 2 1 + x 2 − x 5 ( 1 + x 2 ) 3 = x ( 3 + x 2 ) 2 ( 1 + x 2 ) 3 {\displaystyle {\frac {dy}{dx}}={\frac {3{\sqrt {x}}}{2{\sqrt {1+x^{2}}}}}-{\frac {\sqrt {x^{5}}}{\sqrt {(1+x^{2})^{3}}}}={\frac {{\sqrt {x}}(3+x^{2})}{2{\sqrt {(1+x^{2})^{3}}}}}} .
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![{\displaystyle {\frac {dy}{dx}}={\tfrac {3}{2}}x^{\tfrac {1}{2}}(1+x^{2})^{-{\tfrac {1}{2}}}+x^{\tfrac {3}{2}}\times {\frac {d\left[(1+x^{2})^{-{\tfrac {1}{2}}}\right]}{dx}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53d425805fe3a8e85201706468dd6d0c6f01298a)
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, as shown in example (2) above, we get
![{\displaystyle {\frac {d\left[(1+x^{2})^{-{\tfrac {1}{2}}}\right]}{dx}}=-{\frac {x}{\sqrt {(1+x^{2})^{3}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5331a654a85891bf74fe2eec26d29e8ee5a76eb7)
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