Total Derivatives Engineering Maths, Btech first year

Total Derivatives

If there is a function $u$, $u=f(x,y)$ where $x=\mathrm{\Phi }(t)$ and $y=\mathrm{\Psi }(t)$, 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 $x=a{t}^2$, $y=2at$ 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 $$\frac{du}{dt}=4a(a{t}^2)(t)+4a(2at)$$ $$\frac{du}{dt}=4a^{2}t(t^2+2)$$

Sample Problem- Implicit Function

If $z=x^{2}y$ and $x^2+xy+y^2=1$ find $dz/dx$

Solution :

$$z=xy⟶(1)$$ $$x^2+xy+y^2=1⟶(2)$$ 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) $$\frac{dz}{dx}=2xy-(x^2)\frac{(2x+y)}{(x+2y)}$$ $$\boxed{{\displaystyle \frac{dz}{dx}=\frac{x^{2}y+4x{y}^2-2x^3}{x+2y}}}$$