**Partial Derivative of functions **is an important topic in Calculus. If we have a function f(x,y) i.e. a function which depends on two variables x and y, where x and y are independent to each other, then we say that the function f partially depends on x and y. The derivative of f is called the partial derivative of f. When we differentiate f with respect to x, then consider y as a constant and when we differentiate f with respect to y, then consider x as a constant.

*For example, *suppose f is a function in x and y then it will be denoted by f(x,y). So, partial derivative of f with respect to x will be

It can also be denoted as

**Partial Derivative Identities**

There are some identities for partial derivatives as per the definition of the function.

1. If u = f(x,y) and both x and y are differentiable of t i.e. x = g(t) and y = h(t), then the term differentiation becomes total differentiation.

2. The total partial derivative of u with respect to t is

3. If f is a function defined as f(x), where x(u,v) then

Suppose f = f(x,y) and y is a implicit function it means y is itself a function of x, then

4. If f(x,y) where x(u,v) and y(u,v), then

**Partial Derivative Formula**

If f(x,y) is a function, where f is partially depends on x and y. Then if we differentiate f withe respect to x and y then the derivatives are called the partial derivative of f with respect to x and y. The formula for partial derivative of f with respect to x taking y as a constant,

And partial derivative of f with respect to y taking x as a constant,

**First Partial Derivative**

If u = f(x,y) is then the partial derivative of f with respect to x defined as

And, partial derivative of f with respect to y defined as

These are called the first partial derivatives of f. When we calculate the partial derivatives of f with respect to x treating y as a constant and vise versa.

**Double Partial Derivative**

Since the second order partial derivative can be found by differentiating the first partial derivative, we can also call it as the double partial derivative. Second order partial derivatives can be defined as follows:

The partial differentiation f_{xy} and f_{yx} are distinguished by the order on which ‘f’ is successively differentiated with respect to ‘x’ and ‘y’. In general, the two partial derivatives f_{xy} and f_{yx} need not be equal.

**Second Partial Derivative Test**

The necessary condition for the existence of relative maximum and relative minimum of a function of two variables f(x,y) is

If (x_{1} , y_{1}) are the points of the function which satisfying equation (a), then

If f_{xx} < 0 and f_{yy} < 0 then (x_{1} , y_{1}) is the relative maximum point of the function.

If fxx < 0 and fyy < 0 then (x_{1} , y_{1}) is the relative minimum point of the function.

**Mixed Partial Derivative**

We can find out the mixed partial derivative or cross partial derivative of any function when the second order partial derivative exists. If f(x,y) is a function of with two independent variables, then we know that

Then,

The terms

**Higher-Order Partial Derivatives**

Second and higher-order partial derivatives are defined analogously to the higher-order derivatives of univariate functions. For the function f (x, y, . . .) the “own” second partial derivative with respect to x is simply the partial derivative of the partial derivative (both with respect to x).

The four second-order partial derivatives are as follows.

Let f (x, y) be a function of two variables such that

1] The partial derivative of

2] The partial derivative of

3] The partial derivative of

4] The partial derivative of

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**Solved Examples**

**Question 1: **Consider the function f(x,y) = 5x^{4}y^{2}+6x^{2}y^{3 }. Find f_{x} and f_{y}.

**Solution:**

Given f(x,y) = 5x^{4}y^{2}+6x^{2}y^{3}

Take y as constant and differentiate the given function w.r.t x to find f_{x}

f_{x} = 20x^{3}y^{2}+12xy^{3}

Take x as constant and differentiate the given function w.r.t y to find fy

f_{y} = 5x^{4}(2y)+6x^{2}(3y^{2}) = 10x^{4}y+18x^{2}y^{2}

**Question 2:** Find the partial derivatives f_{xy} and f_{yx} of the function f(x,y) =

**Solution:**

To find fxy and fyx first we have to find fx and fy. So, ‘y’ is held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’

Or

Now, fxy =

= ^{2}y^{4} – y cos x]

= 12x^{2}y^{3} – cos x

Again, f_{yx} =

= ^{3}y^{3} – sin x]

= 12x^{2}y^{3} – cos x

**Question 3:** If f(x,y) = xy and y = e^{x}, find

**Solution:**

In this function, y is an implicit function, then we use

Now, ^{x}

Then, ^{x }= e^{x }+ x e^{x} = e^{x }(1 + x)

**Question 4:** If f = e^{ax} sin (by), where ‘a’ and ‘b’ are real constants. Find f_{xx} , f_{yy} , f_{xy} , f_{yx}

**Solution:**

Given, f = e^{ax }sin (by),

**(i) **f_{xx} = (f_{x})_{x}

To find f_{x} , ‘y’ is held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’

f_{xx} = [a. e^{ax} sin (by)]x

To find f_{xx} , once again ‘y’ is held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’

f_{xx} = a^{2}. e^{ax} sin (by)

**(ii)** f_{yy} = (f_{y})_{y}

To find fy , ‘x’ is held constant and the resulting function of ‘y’ is differentiated with respect to ‘y’

f_{yy} = [b. e^{ax} cos (by)] y

To find f_{yy} , once again ‘x’ is held constant and the resulting function of ‘y’ is differentiated with respect to ‘y’

f_{yy} = – b^{2}. e^{ax} sin (by)

**(iii)** f_{xy} = (f_{x})_{y}

To find f_{x} , ‘y’ is held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’

f_{xy} = [a. e^{ax} sin (by)] y

fxy = ab. e^{ax} cos (by)

**(iv)** f_{yx} = (f_{y})_{x}

f_{yx} = [b. e^{ax} cos (by)] x

f_{yx} = ab. e^{ax} cos (by)

**Question 5: **If

**Solution: **

**Question 6:** If

**Solution:**

**Question 7: **If

**Solution:**