Question

Find the first partial derivatives of the following functions. g(x,y)=y\sin^{-1}\sqrt{xy}

Derivatives
ANSWERED
asked 2021-03-23
Find the first partial derivatives of the following functions.
\(\displaystyle{g{{\left({x},{y}\right)}}}={y}{{\sin}^{{-{1}}}\sqrt{{{x}{y}}}}\)

Answers (1)

2021-03-25
Step 1
We know that first partial derivatives is given by \(\displaystyle{\frac{{\partial{g{{\left({x},{y}\right)}}}}}{{\partial{x}}}}{\quad\text{and}\quad}{\frac{{\partial{g{{\left({x},{y}\right)}}}}}{{\partial{y}}}}\) for the given function.
Here we will partial differentiate the given equation w.r.t x and y.
Step 2
According to question
\(\displaystyle{g{{\left({x},{y}\right)}}}={y}{{\sin}^{{-{1}}}{\left(\sqrt{{{x}{y}}}\right)}}\)
Applying partial derivative rule,
First partial derivative w.r.t ' x' is given by
\(\displaystyle{\frac{{\partial{g{{\left({x},{y}\right)}}}}}{{\partial{x}}}}={y}{\frac{{{1}}}{{\sqrt{{{1}-{x}{y}}}}}}{\left({\frac{{{1}}}{{{2}\sqrt{{{x}{y}}}}}}\right)}\)
First partial derivative w.r.t 'y' is given by
\(\displaystyle{\frac{{\partial{g{{\left({x},{y}\right)}}}}}{{\partial{y}}}}={y}{\frac{{{1}}}{{\sqrt{{{1}-{x}{y}}}}}}{\left({\frac{{{1}}}{{{2}\sqrt{{{x}{y}}}}}}\right)}+{{\sin}^{{-{1}}}{\left(\sqrt{{{x}{y}}}\right)}}\)
0
 
Best answer

expert advice

Need a better answer?
...