Total Derivatives
If there is a function \(u\), \(u = f(x,y)\) where \(x = \Phi (t) \) and \(y = \Psi (t) \), then \(u\) can be expressed as a function of \(t\) by substituting values of \(x\) and \(y\) in \(f(x,y)\) and hence, we can find \(du/dt\) which is called as the Total Derivative of \(u\).
There are two methods to find total derivative
In Substitution, as the name suggests, we substitute values of \(x\) and \(y\) in \(u\) but this method fails if the function given is an implicit function(\(y\) is dependent on \(x\) ).
To find derivative without substitution, we have something called Chain Rule which is,
\[ \frac{du}{dt} = \frac{\partial u}{\partial x}\times \frac{dx}{dt} + \frac{\partial u}{\partial y}\times \frac{dy}{dt} \]
If \(u = f(x,y,z) \) where \(x,y,z\) all are functions of variable \(t\), then Chain Rule is,
\[ \frac{du}{dt} = \frac{\partial u}{\partial x}\times \frac{dx}{dt} + \frac{\partial u}{\partial y}\times \frac{dy}{dt} + \frac{\partial u}{\partial z}\times \frac{dz}{dt} \]
It's easy to remember if you understand this- \(u\) is a function of \(x, y\), \(x, y\) are functions of \(t\), and we have to reach from \(u\) to \(t\), as shown in the figure,
Now, read this carefully: if a function is of only one variable, then Partial Derivative and regular differentiation, both give same results(you can try yourself). Here those functions are \(x\) and \(y\), thus \(dx/dt\).
If a function has more than one variable, then only P.D. is possible, here that function is \(u\), therefore, \(\partial u/\partial x\). So if carefully observed, Chain Rule is just a combination of P.D. and differentiation, both depending on number of variables a function has.
Sample Problem :
If \( u = x^2 + y^2 \) and \( x = at^2 \), \( y = 2at \) find \( du/dt \)
Solution :
By Total Differentiation, \[ \frac{du}{dt} = \frac{\partial u}{\partial x}\times \frac{dx}{dt} + \frac{\partial u}{\partial y}\times \frac{dy}{dt} \] \[ \frac{du}{dt} = (2x + 0)\times (2at) + (0 + 2y)\times (2a) \] \[ \frac{du}{dt} = 4axt + 4ay \] But the answer must be in that variable w.r.t. which we are calculating total derivative \[ \frac{du}{dt} = 4a(at^2)(t) + 4a(2at) \] \[ \frac{du}{dt} = 4a^2t(t^2 + 2) \]
Sample Problem- Implicit Function
If \( z = x^2y \) and \( x^2 + xy + y^2 = 1 \) find \( dz/dx \)
Solution :
\[ z = xy \hspace{20pt} \longrightarrow (1) \] \[ x^2 + xy + y^2 = 1 \hspace{20pt} \longrightarrow (2) \] Now here, \(x\) is the independent variable and \(y\) is dependent on \(x\) (there is no \(t\) here), therefore, \[ \frac{dz}{dx} = + \frac{\partial z}{\partial y}\times \frac{dy}{dx} \] Partially Differentiating equation (1) w.r.t. \(x\) and \(y\) \[ \frac{\partial z}{\partial x} = 2xy \] \[ \frac{\partial z}{\partial y} = x^2 \] Differentiating equation (2) w.r.t. \(x\) \[ 2x + x\frac{dy}{dt} + y + 2y\frac{dy}{dx} = 0 \] \[ \Rightarrow \frac{dy}{dx} = -\frac{(2x + y)}{(x + 2y)} \] Put values of derivatives in equation (3) \[ \frac{dz}{dx} = 2xy - (x^2)\frac{(2x + y)}{(x + 2y)} \] \[ \boxed{ \frac{dz}{dx} = \frac{x^2y + 4xy^2 - 2x^3}{x + 2y} } \]
