
What is the derivative of \[\sqrt{xy}\]?
Answer
516.3k+ views
Hint: In this problem, we have to find the derivative of \[\sqrt{xy}\], where we have two variables and we can use the partial differentiation. We should know that partial differentiation is used to differentiate mathematical functions having more than one variable in them. We can use the partial differentiation, as it holds some independent variable as constant and find the derivative with respect to another independent variable.
Complete step by step answer:
We know that partial differentiation is used to differentiate mathematical functions having more than one variable in them.
We already know that in ordinary differentiation we find the derivative with respect to one variable only, as the function contains only one variable. But if we have more than one variable for differentiation, then we should use the term ‘partial differentiation’, to differentiate the given function. It’s symbol is denoted by \[\partial \].
We can use the partial differentiation, as it holds some independent variable as constant and find the derivative with respect to another independent variable.
We know that the given function is,
\[f\left( x,y \right)=\sqrt{xy}\]
We can now write it as,
\[f\left( x,y \right)={{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}\]
We can see that here we have two variables.
We can now find the partial derivative with respect to y, we consider x as constant.
Partial derivative with respect to y is,
\[\Rightarrow f{{'}_{y}}=\dfrac{\partial f}{\partial y}={{x}^{\dfrac{1}{2}}}\times \dfrac{1}{2}{{y}^{-\dfrac{1}{2}}}=\left( \dfrac{1}{2} \right){{\left( \dfrac{x}{y} \right)}^{\dfrac{1}{2}}}\]
We can now find the partial derivative with respect to x, we consider y as constant.
Partial derivative with respect to x is,
\[\Rightarrow f{{'}_{x}}=\dfrac{\partial f}{\partial x}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}}\times {{y}^{\dfrac{1}{2}}}=\left( \dfrac{1}{2} \right){{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{2}}}\].
Therefore, Partial derivative with respect to x is,
\[\Rightarrow f{{'}_{x}}=\left( \dfrac{1}{2} \right){{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{2}}}\].
Partial derivative with respect to y is,
\[\Rightarrow f{{'}_{y}}=\left( \dfrac{1}{2} \right){{\left( \dfrac{x}{y} \right)}^{\dfrac{1}{2}}}\]
Note: We should remember that if we have three or more variables, we will differentiate with respect to one variable and we will consider all other variables as constants. We should also remember that partial differentiation can be done only if ordinary differentiation rules and formulas are known.
Complete step by step answer:
We know that partial differentiation is used to differentiate mathematical functions having more than one variable in them.
We already know that in ordinary differentiation we find the derivative with respect to one variable only, as the function contains only one variable. But if we have more than one variable for differentiation, then we should use the term ‘partial differentiation’, to differentiate the given function. It’s symbol is denoted by \[\partial \].
We can use the partial differentiation, as it holds some independent variable as constant and find the derivative with respect to another independent variable.
We know that the given function is,
\[f\left( x,y \right)=\sqrt{xy}\]
We can now write it as,
\[f\left( x,y \right)={{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}\]
We can see that here we have two variables.
We can now find the partial derivative with respect to y, we consider x as constant.
Partial derivative with respect to y is,
\[\Rightarrow f{{'}_{y}}=\dfrac{\partial f}{\partial y}={{x}^{\dfrac{1}{2}}}\times \dfrac{1}{2}{{y}^{-\dfrac{1}{2}}}=\left( \dfrac{1}{2} \right){{\left( \dfrac{x}{y} \right)}^{\dfrac{1}{2}}}\]
We can now find the partial derivative with respect to x, we consider y as constant.
Partial derivative with respect to x is,
\[\Rightarrow f{{'}_{x}}=\dfrac{\partial f}{\partial x}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}}\times {{y}^{\dfrac{1}{2}}}=\left( \dfrac{1}{2} \right){{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{2}}}\].
Therefore, Partial derivative with respect to x is,
\[\Rightarrow f{{'}_{x}}=\left( \dfrac{1}{2} \right){{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{2}}}\].
Partial derivative with respect to y is,
\[\Rightarrow f{{'}_{y}}=\left( \dfrac{1}{2} \right){{\left( \dfrac{x}{y} \right)}^{\dfrac{1}{2}}}\]
Note: We should remember that if we have three or more variables, we will differentiate with respect to one variable and we will consider all other variables as constants. We should also remember that partial differentiation can be done only if ordinary differentiation rules and formulas are known.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw a ray diagram of compound microscope when the class 12 physics CBSE

How is democracy better than other forms of government class 12 social science CBSE

What is virtual and erect image ?

Explain the energy losses in the transformer How are class 12 physics CBSE

