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

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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Arnold Odonnell
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
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{d}{w}}}}={\frac{{{d}}}{{{d}{w}}}}{\left[{\left({w}^{{{2}}}-{1}\right)}^{{-{\frac{{{1}}}{{{2}}}}}}\right]}$$
$$\displaystyle=-{\frac{{{1}}}{{{2}}}}{\left({w}^{{{2}}}-{1}\right)}^{{-{\frac{{{1}}}{{{2}}}}-{1}}}\cdot{\frac{{{d}{\left({w}^{{{2}}}-{1}\right)}}}{{{d}{w}}}}\ \ \ {\left[\because{\frac{{{d}{\left({x}^{{{n}}}\right)}}}{{{\left.{d}{x}\right.}}}}={n}{x}^{{{n}-{1}}}\right]}$$
$$\displaystyle=-{\frac{{{1}}}{{{2}}}}{\left({w}^{{{2}}}-{1}\right)}^{{-{\frac{{{3}}}{{{2}}}}}}\cdot{\left({2}{w}-{0}\right)}$$
$$\displaystyle=-{\frac{{{w}}}{{{\left({w}^{{{2}}}-{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}}}}$$