Derivative of a Vector Function Partial Differentiation
So far the reader has studied the differentiation of vector functions in one variable. However, a vector may be a function of more than one scalar variable.
Let f be the vector function of scalar variables p, q, t over a domain S, then we write f = f(p, q, t).
Treating t as the variable and p, q as constants; f can be considered as the vector function of the scalar variable t over a domain S.
⇒ \(\text { If Lt } \mathrm{L}_{\delta t \rightarrow 0} \frac{\mathrm{I}(p, q, t+\delta t)-\mathrm{f}(p, q, t)}{\delta \mathrm{q}}\)
exists then f is said to have partial derivative w.r.to ’ t ’ and is denoted by \(\frac{\partial \mathrm{O}^{\delta_2}}{\partial \mathrm{I}}\)
Similarly, treating q as the variable and taking p, and t as constants
⇒ \(\frac{\partial \mathrm{f}}{\partial q}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}(p, q+\delta q, t)-\mathrm{f}(p, q, t)}{\delta q} \text { etc. }\)
Let A, B, φ be the functions of more than one scalar variable. Then the following can be verified.
1. \(\frac{\partial}{\partial t}(\phi \mathbf{A})=\frac{\partial \phi}{\partial t} \mathbf{A}+\phi \frac{\partial \mathbf{A}}{\partial t}\)
2. If λ is a constant then \(\frac{\partial}{\partial t}(\lambda \mathbf{A})=\lambda \frac{\partial \mathbf{A}}{\partial t}\)
3. If c is a constant vector then \(\frac{\partial}{\pi}(\phi \mathbf{c})=\frac{\partial \phi}{\partial t} \mathbf{c}\)
4. \(\frac{\partial}{\partial t}(\mathbf{A} \pm \mathbf{B})=\frac{\partial \mathbf{A}}{\partial t} \pm \frac{\partial \mathrm{B}}{\partial t} 5 \cdot \frac{\partial}{\partial t}(\mathbf{A} \cdot \mathbf{B})=\frac{\partial \mathrm{A}}{\partial t} \cdot \mathbf{B}+\mathbf{A} \cdot \frac{\partial \mathrm{B}}{\partial t}\)
5. \(\frac{\partial}{\partial t}(\mathbf{A} \times \mathbf{B})=\frac{\partial \mathbf{A}}{\partial t} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathrm{B}}{\partial t}\)
Let f = f1 i + f2 j + f3 k where f1, f2, f3 are differentiable scalar functions of more than one variable t. Then
⇒ \(\frac{\mu_i}{\partial t}=\mathrm{i} \frac{\partial_1}{\partial t}+\mathrm{j} \frac{\partial_2}{\partial t}+\mathrm{k} \frac{\partial_2}{\partial t}\)
Higher partial derivatives are derivatives defined as in the Calculus of real variables. Thus, for instant
⇒ \(\frac{\partial^2 \mathrm{r}}{\partial t^2}=\frac{\partial}{\partial t}\left(\frac{\partial \mathrm{f}}{\partial t}\right), \frac{\partial^2 \mathrm{r}}{\partial t \partial_P}=\frac{\partial}{\partial t}\left(\frac{\partial \mathrm{r}}{\partial \mathrm{p}}\right) \text { etc. }\)
Derivative Of A Vector Function With Scalar Multiplication
Derivative of a Vector Function Solved Problems
Example 1. If \(\mathbf{f}=\left(2 x^2 y-x^4\right) \mathbf{i}+\left(e^{x y}-y \sin x\right) \mathbf{j}+\left(x^2 \cos y\right) \mathbf{k} \text {. find } \frac{\partial^2 f}{\partial x^2} \text { and } \frac{\partial^2 f}{\partial x \partial_y}\)
Solution. Given f = (2x²y − x4) i + (exy − y sin x) j + (x2 cos y) k
⇒ \(\frac{\partial \mathrm{f}}{\partial x}\)= (4xy − 4x3)i + (y exy − y cos x) j + (2x cos y)k
⇒ \(\frac{\partial^2 \mathrm{f}}{\partial x^2}\)= (4y − 12x2)i + (y2 exy + y sin x) j + (2cos y)k
⇒ \(\frac{\partial^2 \mathrm{f}}{x \partial y y}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=4 x \mathbf{i}+\left(e^{x y}+y^2 e^{x y}-\cos x\right) \mathbf{j}-2 x \sin , y \mathbf{k} \)
Example. 2. If A = 2x2 i − 3yzj + xz2k φ = 2z − x3y then find
⇒ \(\mathbf{A} \cdot\left[\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right] \text { and } \mathbf{A} \times\left[\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right] \text { at }(1,-1,1)\)
Solution:
Given
A = 2x2 i − 3yzj + xz2k φ = 2z − x3y
\(\phi=2 z-x^3 y . \quad ∴ \frac{\partial \phi}{\partial x}=-3 x^2 y, \frac{\partial \phi}{\partial y}=-x^3, \frac{\partial \phi}{\partial z}=2 \)At(1,-1,1), \(\mathbf{A}=2 \mathbf{i}+3 \mathbf{j}+\mathbf{k} \text { and } \frac{\partial \phi}{\partial x}=+3, \frac{\partial \phi}{\partial y}=-1, \frac{\partial \phi}{\partial z}=2\)
⇒ \(\mathbf{A} \cdot\left[\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right]_{(1,-1,1)}=(2 \mathbf{i}+3 \mathbf{j}+\mathbf{k}) \cdot(+3 \mathbf{i}-\mathbf{j}+2 \mathbf{k})=+6-3+2=5\)
and \(\mathbf{A} \times\left[\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right]=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 1 \\
3 & -1 & 2
\end{array}\right|=7 \mathbf{i}-\mathbf{j}-11 \mathbf{k}\)