Question

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

Derivatives
Find both first partial derivatives. $$\displaystyle{z}={\sin{{\left({x}+{2}{y}\right)}}}$$

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.
$$\displaystyle{\frac{{\partial{z}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\sin{{\left({x}+{2}{y}\right)}}}$$
$$\displaystyle={\left[{\cos{{\left({x}+{2}{y}\right)}}}\right]}{\frac{{\partial{z}}}{{\partial{y}}}}{\left({x}+{2}{y}\right)}$$
$$\displaystyle={\left[{\cos{{\left({x}+{2}{y}\right)}}}\right]}{\left({2}\right)}$$
$$\displaystyle={2}{\cos{{\left({x}+{2}{y}\right)}}}$$