Consider a function z = xy+x(y^2+1).Find first order partial derivatives, total differential, and total derivative with respect to x.

Consider a function z = xy+x(y^2+1).Find first order partial derivatives, total differential, and total derivative with respect to x.

Question
Derivatives
asked 2021-02-09
Consider a function \(\displaystyle{z}={x}{y}+{x}{\left({y}^{{2}}+{1}\right)}\).Find first order partial derivatives, total differential, and total derivative with respect to x.

Answers (1)

2021-02-10
Step 1
The function is \(\displaystyle{z}={x}{y}+{x}{\left({y}^{{2}}+{1}\right)}\)
find the partial derivatives
\(\displaystyle\frac{{\partial{z}}}{{\partial{x}}}={y}+{\left({y}^{{2}}+{1}\right)}\)
\(\displaystyle\frac{{\partial{z}}}{{\partial{x}}}={x}+{x}{\left({2}{y}\right)}\)
The partial derivatives are:
\(\displaystyle\frac{{\partial{z}}}{{\partial{x}}}={y}+{y}^{{2}}+{1}\)
\(\displaystyle\frac{{\partial{z}}}{{\partial{x}}}={x}+{2}{x}{y}\)
Step 2
total differential is given by \(\displaystyle{\left.{d}{z}\right.}=\frac{{\partial{z}}}{{\partial{x}}}{\left.{d}{x}\right.}+\frac{{\partial{z}}}{{\partial{y}}}{\left.{d}{y}\right.}\)
Substitute the values
\(\displaystyle{\left.{d}{z}\right.}={\left({y}+{y}^{{2}}+{1}\right)}{\left.{d}{x}\right.}+{\left({x}+{2}{x}{y}\right)}{\left.{d}{y}\right.}\)
The total differential is:
\(\displaystyle{\left.{d}{z}\right.}={\left({y}+{y}^{{2}}+{1}\right)}{\left.{d}{x}\right.}+{\left({x}+{2}{x}{y}\right)}{\left.{d}{y}\right.}\)
Step 3
The total derivative with respect to x is given by
\(\displaystyle\frac{{\partial{z}}}{{\partial{x}}}=\frac{{\partial{z}}}{{\partial{x}}}+\frac{{\partial{z}}}{{\partial{y}}}\frac{{\partial{y}}}{{\partial{x}}}\)
Substitute the values
\(\displaystyle\frac{{\partial{z}}}{{\partial{x}}}={\left({y}+{y}^{{2}}+{1}\right)}+{\left({x}+{2}{x}{y}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\)
The total derivative with respect to x is:
\(\displaystyle\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{x}\right.}}}={\left({y}+{y}^{{2}}+{1}\right)}+{\left({x}+{2}{x}{y}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\)
0

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