Question

Find both first partial derivatives. z=\sin(x+2y)

Derivatives
ANSWERED
asked 2021-02-24
Find both first partial derivatives. \(\displaystyle{z}={\sin{{\left({x}+{2}{y}\right)}}}\)

Expert Answers (1)

2021-02-26
Step 1
The function is \(\displaystyle{z}={\sin{{\left({x}+{2}{y}\right)}}}\)
Determine the partial derivative with respect to x.
\(\displaystyle{\frac{{\partial{z}}}{{\partial{x}}}}={\frac{{\partial}}{{\partial{x}}}}{\sin{{\left({x}+{2}{y}\right)}}}\)
\(\displaystyle={\left[{\cos{{\left({x}+{2}{y}\right)}}}\right]}{\frac{{\partial{z}}}{{\partial{y}}}}{\left({x}+{2}{y}\right)}\)
\(\displaystyle={\cos{{\left({x}+{2}{y}\right)}}}\)
Step 2
Determine the partial derivative with respect to y.
\(\displaystyle{\frac{{\partial{z}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\sin{{\left({x}+{2}{y}\right)}}}\)
\(\displaystyle={\left[{\cos{{\left({x}+{2}{y}\right)}}}\right]}{\frac{{\partial{z}}}{{\partial{y}}}}{\left({x}+{2}{y}\right)}\)
\(\displaystyle={\left[{\cos{{\left({x}+{2}{y}\right)}}}\right]}{\left({2}\right)}\)
\(\displaystyle={2}{\cos{{\left({x}+{2}{y}\right)}}}\)
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