Partial Differentiation

Partial Derivatives :

Suppose z = f(x , y) be a function of two independent variables x and y , if we keep y as a constant and x varies , then z becomes a function of x only . The derivative of z with respect to x , keeping y as a constant is called partial derivative of ‘ z ‘ w.t.t ‘ x ‘ and it is denoted by :

$\large \frac{\partial z}{\partial x} \; , \; \frac{\partial f}{\partial x}$ etc.

Similarly , we can write partial derivative of ‘ z ‘ w.r.t ‘ y ‘ keeping x as a constant is denoted by :

$\large \frac{\partial z}{\partial y} \; , \; \frac{\partial f}{\partial y}$ etc.

Solved Example : If $\large z = \frac{x^3 + y^3}{x + y}$ ; find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$

Solution : $\large z = \frac{x^3 + y^3}{x + y}$

Using Quotient Rule ;

$\large \frac{\partial z}{\partial x} = \frac{(x + y) 3 x^2 – (x^3 + y^3) .1 }{(x + y)^2}$

$\large \frac{\partial z}{\partial x} = \frac{3 x^3 + 3 x^2 y – x^3 – y^3 }{(x + y)^2}$

$\large \frac{\partial z}{\partial x} = \frac{2 x^3 + 3 x^2 y – y^3 }{(x + y)^2}$

Similarly ,

$\large \frac{\partial z}{\partial y} = \frac{(x + y) 3 y^2 – (x^3 + y^3) .1 }{(x + y)^2}$

$\large \frac{\partial z}{\delta y} = \frac{3 x y^2 + 3 y^3 – x^3 – y^3 }{(x + y)^2}$

$\large \frac{\partial z}{\partial y} = \frac{3 x y^2 + 2 y^3 – x^3 }{(x + y)^2}$

Problem : If z = eax+by.f(ax-by) , then Prove that $b \frac{\partial z}{\partial x} + a \frac{\partial z}{\partial y} = 2 a b z$

Solution :  z = eax+by.f(ax-by)

$ \frac{\partial z}{\partial x} = a e^{ax+by} . f(ax-by) + e^{ax+by}.a f'(ax-by)$

Multiplying by b

$ b \frac{\partial z}{\partial x} = a b e^{ax+by} . f(ax-by) + a b e^{ax+by}. f'(ax-by)$ …(i)

$ \frac{\partial z}{\partial y} = b e^{ax+by} . f(ax-by) + e^{ax+by}.(-b) f'(ax-by)$

Multiplying by a

$ a \frac{\partial z}{\partial y} = a b e^{ax+by} . f(ax-by) – a b e^{ax+by}. f'(ax-by)$ …(ii)

Now adding (i) & (ii)

$ b \frac{\partial z}{\partial x} + a \frac{\partial z}{\partial y} = 2 a b e^{ax+by} . f(ax-by) $

$ b \frac{\partial z}{\partial x} + a \frac{\partial z}{\partial y} = 2 a b z$

Problem : Prove that y = f(x + at) + g(x – at) satisfies $ \frac{\partial^2 y}{\partial t^2} = a^2 (\frac{\partial^2 y}{\partial x^2})$ ; Where f and g are assumed to be twice differentiable and a is any constant .

Solution: y = f(x + at) + g(x – at) ….(i)

Differentiable partially (i) with respect to x ;

$\frac{\partial y}{\partial x} = f'(x + at) + g'(x – at) $

$\frac{\partial^2 y}{\partial x^2} = f”(x + at) + g”(x – at) $

Differentiable partially (i) with respect to t ;

$\frac{\partial y}{\partial t} = f'(x + at) . a + g'(x – at) (-a)$

$\frac{\partial^2 y}{\partial t^2} = a^2 f”(x + at) + a^2 g”(x – at) $

= a2[f”(x + at) + g”(x – at)

$ = a^2 \frac{\partial^2 y}{\partial x^2}$ Proved

Problem : If u = exyz ; find the value of $\frac{\partial^3 u}{\partial x \partial y \partial z }$

Solution : u = exyz

$\frac{\partial u}{\partial z} = e^{x y z}(x y) $

$\frac{\partial^2 u}{\partial y \partial z} = e^{x y z}(x) + e^{x y z}(x z)(x y) $

= exyz (x + x2 y z )

$\frac{\partial^3 u}{\partial x \partial y \partial z} = e^{x y z}(1 + 2 x y z) + e^{x y z}(y z)(x + x^2 y z) $

$= e^{x y z} [1 + 2 x y z + x y z + x^2 y^2 z^2 ] $

$ = e^{x y z} [1 + 3 x y z + x^2 y^2 z^2 ] $

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