Find the indicated derivatives. \frac{dz}{dw}\ if\ z=\frac{1}{\sqrt{w^{2}

Globokim8 2021-10-02 Answered
Find the indicated derivatives.
\(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{d}{w}}}}\ {\quad\text{if}\quad}\ {z}={\frac{{{1}}}{{\sqrt{{{w}^{{{2}}}-{1}}}}}}\)

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Expert Answer

Arnold Odonnell
Answered 2021-10-03 Author has 8362 answers
Given,
\(\displaystyle{z}={\frac{{{1}}}{{\sqrt{{{w}^{{{2}}}-{1}}}}}}\)
\(\displaystyle\Rightarrow{z}={\frac{{{1}}}{{{\left({w}^{{{2}}}-{1}\right)}^{{{\frac{{{1}}}{{{2}}}}}}}}}\)
\(\displaystyle\Rightarrow{z}={\left({w}^{{{2}}}-{1}\right)}^{{-{\frac{{{1}}}{{{2}}}}}}\)
Step 2
Now differentiating with respect to w, we get
\(\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{d}{w}}}}={\frac{{{d}}}{{{d}{w}}}}{\left[{\left({w}^{{{2}}}-{1}\right)}^{{-{\frac{{{1}}}{{{2}}}}}}\right]}\)
\(\displaystyle=-{\frac{{{1}}}{{{2}}}}{\left({w}^{{{2}}}-{1}\right)}^{{-{\frac{{{1}}}{{{2}}}}-{1}}}\cdot{\frac{{{d}{\left({w}^{{{2}}}-{1}\right)}}}{{{d}{w}}}}\ \ \ {\left[\because{\frac{{{d}{\left({x}^{{{n}}}\right)}}}{{{\left.{d}{x}\right.}}}}={n}{x}^{{{n}-{1}}}\right]}\)
\(\displaystyle=-{\frac{{{1}}}{{{2}}}}{\left({w}^{{{2}}}-{1}\right)}^{{-{\frac{{{3}}}{{{2}}}}}}\cdot{\left({2}{w}-{0}\right)}\)
\(\displaystyle=-{\frac{{{w}}}{{{\left({w}^{{{2}}}-{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}}}}\)
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