Question

$\mathbf{If}\mathbf{}\mathit{m}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathbf{}\mathbf{and}\mathbf{}\mathit{n}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{}\mathbf{find}\mathbf{}{\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{b}}{\mathbf{\partial }\mathbf{m}}\mathbf{\right)}}_{\mathbf{n}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{m}}{\mathbf{\partial }\mathbf{b}}\mathbf{\right)}}_{\mathbf{a}}$

Short answer

The value of ${\left(\frac{\partial b}{\partial m}\right)}_{\mathrm{n}}\mathrm{is}\frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}$.

The value of ${\left(\frac{\partial m}{\partial b}\right)}_{\mathrm{n}}is1$.

Step-by-step solution

Given information.

Given$m=a+b,\mathrm{n}=a^2+b^2$,

Definition of partial differentiation.

Partial differentiation is defined as the process, in which find the partial derivative of a function.

In Partial differentiation, the function has more than one variable and find the partial derivative of a function with respect to one variable and keeping the other variable constant.

Find the partial differentiation of each functions.

Now find the partial differentiation of function,$m=a+b$ with respect to m.

$1=\frac{\partial a}{\partial m}+\frac{\partial b}{\partial m}.........\left(1\right)$

Now find the partial differentiation of function, with respect to.

$$\begin{gathered} \frac{\partial n}{\partial m}=2a\frac{\partial a}{\partial m}+2b\frac{\partial b}{\partial m} \\ 2a\frac{\partial a}{\partial m}=2a\frac{\partial n}{\partial m}+2b\frac{\partial b}{\partial m} \\ \frac{\partial a}{\partial m}=\frac{1}{2a}\left(\frac{\partial n}{\partial m}+2b\frac{\partial b}{\partial m}\right)...........\left(2\right) \end{gathered}$$

Substitute $\frac{\partial a}{\partial m}$from equation (2) to equation (1).

role="math" localid="1660821307849" $1=\frac{1}{2a}\left(\frac{\partial n}{\partial m}-2b\frac{\partial b}{\partial m}\right)+\frac{\partial b}{\partial m}$

Find (∂b∂m)n.

Now find ${\left(\frac{\partial b}{\partial m}\right)}_n$by keeping n is a constant.

$$\begin{gathered} 1=\frac{1}{2a}\left(0-2b{\left(\frac{\partial b}{\partial m}\right)}_n\right)+{\left(\frac{\partial b}{\partial m}\right)}_n \\ 1=-\frac{b}{a}.{\left(\frac{\partial b}{\partial m}\right)}_n+{\left(\frac{\partial b}{\partial m}\right)}_n \\ {\left(\frac{\partial b}{\partial m}\right)}_n\left(1-\frac{b}{a}\right)=1 \\ {\left(\frac{\partial b}{\partial m}\right)}_n=\frac{1}{\left(1-\frac{b}{a}\right)} \\ =\frac{a}{a-b} \end{gathered}$$

Hence the value of${\left(\frac{\partial b}{\partial m}\right)}_n\mathrm{is}\frac{\mathrm{a}}{\mathrm{a}-\mathrm{b}}$ .

Find (∂m∂b)n.

Now find the partial differentiation of function,$m=a+b$ with respect to b.

$\frac{\partial m}{\partial b}=\frac{\partial a}{\partial b}+1$

Now find${\left(\frac{\partial m}{\partial b}\right)}_a$by keeping a is constant.

$$\begin{gathered} {\left(\frac{\partial m}{\partial b}\right)}_a=0+1 \\ {\left(\frac{\partial m}{\partial b}\right)}_a=1 \end{gathered}$$

Hence the value of .${\left(\frac{\partial m}{\partial b}\right)}_a\mathrm{is}1$