Question

- If \[u\left( {x,y} \right) = {e^{{x^2} + {y^2}}}\], then find the value of \[\dfrac{{\partial u}}{{\partial x}}\]

Answer 1

- Hint: In this question a function \[u\] is given as a function of two independent variables \[x\] and \[y\]. We have to find the value of \[\dfrac{{\partial u}}{{\partial x}}\]. So, differentiate the function \[u\] partially with respect to \[x\] and keeping \[y\] as constant using the chain rule.

Formula Used:
Chain Rule: If \[u\] is given as a function of \[z\] where \[z\] itself a function of two independent variables \[x\] and \[y\], then \[\dfrac{{\partial u}}{{\partial x}} = \dfrac{{\partial u}}{{\partial z}} \cdot \dfrac{{\partial z}}{{\partial x}}\]
Power Rule: \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], where \[n\] is a real number.
Exponential Rule: \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\]

Complete step-by-step solution:
The given function is \[u\left( {x,y} \right) = {e^{{x^2} + {y^2}}}\]
Let \[z = {x^2} + {y^2}\]
Then \[u = {e^z}\], where \[z\] is a function of two independent variables \[x\] and \[y\]
Using chain rule, we have \[\dfrac{{\partial u}}{{\partial x}} = \dfrac{{\partial u}}{{\partial z}} \cdot \dfrac{{\partial z}}{{\partial x}}.....\left( i \right)\]
So, find the values of \[\dfrac{{\partial u}}{{\partial z}}\] and \[\dfrac{{\partial z}}{{\partial x}}\]
At first let us find the value of \[\dfrac{{\partial u}}{{\partial z}}\]
For a function of one variable, the partial differentiation is the same as the ordinary differentiation.
Here \[u\] is a function of \[z\] only.
So, \[\dfrac{{\partial u}}{{\partial z}} = \dfrac{{du}}{{dz}} = \dfrac{d}{{dz}}\left( {{e^z}} \right) = {e^z}\], using exponential rule
Now, find the value of \[\dfrac{{\partial z}}{{\partial x}}\], where \[z = {x^2} + {y^2}\]
\[z\] is a function of two independent variables \[x\] and \[y\]
So, we have to differentiate the function \[z\] partially with respect to \[x\]
Taking \[y\] as constant, differentiating \[z\] partially with respect to \[x\], we get
\[\dfrac{{\partial z}}{{\partial x}} = \dfrac{\partial }{{\partial x}}\left( {{x^2}} \right) + \dfrac{\partial }{{\partial x}}\left( {{y^2}} \right).....\left( {ii} \right)\]
For a function of one variable, the partial differentiation is same as ordinary differentiation.
So, using the power rule, we have \[\dfrac{\partial }{{\partial x}}\left( {{x^2}} \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right) = 2x\]
and \[\dfrac{\partial }{{\partial x}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {{y^2}} \right) = 0\], since derivative of a constant is zero.
Substituting these expressions in equation \[\left( {ii} \right)\], we get
\[\dfrac{{\partial z}}{{\partial x}} = 2x + 0 = 2x\]
Finally, we get \[\dfrac{{\partial u}}{{\partial z}} = {e^z}\] and \[\dfrac{{\partial z}}{{\partial x}} = 2x\]
Substituting these expressions in equation \[\left( i \right)\], we get
\[\dfrac{{\partial u}}{{\partial x}} = {e^z} \cdot 2x\]
We assumed that \[z = {x^2} + {y^2}\]
So, \[\dfrac{{\partial u}}{{\partial x}} = {e^{{x^2} + {y^2}}} \cdot 2x = 2x{e^{{x^2} + {y^2}}}\]
Hence, the value of \[\dfrac{{\partial u}}{{\partial x}}\] is \[2x{e^{{x^2} + {y^2}}}\]

Note: For a function of two independent variables, you must differentiate the function partially with respect to the variables individually but for a function of one variable, the partial differentiation is same as the ordinary differentiation. Here \[u\] is given as a function of \[z\] where \[z\] itself is a function of two independent variables \[x\] and \[y\], so we use the chain rule \[\dfrac{{\partial u}}{{\partial x}} = \dfrac{{\partial u}}{{\partial z}} \cdot \dfrac{{\partial z}}{{\partial x}}\]