Find the first partial derivatives of the following functions. h(x,y,z)=\cos(x+y+z)

djeljenike 2021-06-03 Answered
Find the first partial derivatives of the following functions.
\(\displaystyle{h}{\left({x},{y},{z}\right)}={\cos{{\left({x}+{y}+{z}\right)}}}\)

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Expert Answer

StrycharzT
Answered 2021-06-04 Author has 12588 answers
Step 1
Consider the given function \(\displaystyle{h}{\left({x},{y},{z}\right)}={\cos{{\left({x}+{y}+{z}\right)}}}\)
Step 2
Now find the partial derivative with respect to x
\(\displaystyle{\frac{{\partial}}{{\partial{x}}}}{\left({h}{\left({x},{y},{z}\right)}\right)}={\frac{{\partial}}{{\partial{x}}}}{\cos{{\left({x}+{y}+{z}\right)}}}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\cdot{\frac{{\partial}}{{\partial{x}}}}{\left({x}+{y}+{z}\right)}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\cdot{\left({1}+{0}+{0}\right)}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\)
Step 3
Now find the partial derivative with respect to y
\(\displaystyle{\frac{{\partial}}{{\partial{y}}}}{\left({h}{\left({x},{y},{z}\right)}\right)}={\frac{{\partial}}{{\partial{x}}}}{\cos{{\left({x}+{y}+{z}\right)}}}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\cdot{\frac{{\partial}}{{\partial{y}}}}{\left({x}+{y}+{z}\right)}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\cdot{\left({0}+{1}+{0}\right)}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\)
Step 4
Now find the partial derivative with respect to z
\(\displaystyle{\frac{{\partial}}{{\partial{z}}}}{\left({h}{\left({x},{y},{z}\right)}\right)}={\frac{{\partial}}{{\partial{x}}}}{\cos{{\left({x}+{y}+{z}\right)}}}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\cdot{\frac{{\partial}}{{\partial{z}}}}{\left({x}+{y}+{z}\right)}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\cdot{\left({0}+{0}+{1}\right)}\)
\(\displaystyle=-{\sin{{\left({x}+{y}+{z}\right)}}}\)
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Answered 2021-11-20 Author has 2252 answers

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