In mathematics, sometimes the function depends on two or more than two variables. In this case, the derivative convert into the partial derivative since the function depends on several variables. Partial derivatives are used in vector calculus and differential geometry.

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If we have a function f(x,y) i.e. a function which is depends on two variable x and y, where x and y are independent to each other. Then we say that the function f is partially depends on x and y. At this time if we calculate the derivative of f, then that derivative is called the partial derivative of f. If we differentiate f with respect to x, then taking y as a constant and if we differentiate f with respect to y, then taking x as a constant.

In mathematics, a**partial derivative **of
a function of several variables is its derivative with respect to one
of those variables, with the others held constant (as opposed to the
total derivative, in which all variables are allowed to vary).

### Partial Derivative Symbol

### Partial Derivative of In

### Partial Derivative Examples

### Solved Examples

**Question 1: **Determine the partial derivative of a function f_{x} and f_{y}: if f(x, y) is given by f(x, y) = tan(xy) + sin x

** Solution: **

**Question 2: **If $f = x^{2}(y – z) + y^{2}(z – x) + z^{2}(x – y)$, prove that $\frac
{\partial f} {\partial x}$ + $\frac {\partial f} {\partial y}$ + $\frac
{\partial f} {\partial z}$$ + 0$

** Solution: **
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,

$f_{x}$ = $\frac{\partial f}{\partial x}$ = $\lim_{h\rightarrow 0}$$\frac{f(x + h,y) - f(x,y)}{h}$

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

$f_{y}$ = $\frac{\partial f}{\partial y}$ = $\lim_{h\rightarrow 0}$$\frac{f(x,y + h) - f(x,y)}{h}$

If u = f(x,y) is then the partial derivative of f with respect to x defined as $\frac{\partial f}{\partial x}$ and denoted by

$\frac{\partial f}{\partial x}$ = $\lim_{\delta x \rightarrow 0}$$\frac{f\left ( x + \delta x , y \right )-f\left ( x,y \right )}{\delta x}$

And, partial derivative of f with respect to y defined as $\frac{\partial f}{\partial y}$ and denoted by

$\frac{\partial f}{\partial y}$ = $\lim_{\delta y\rightarrow 0}$$\frac{f\left ( x,y+\delta y \right )-f\left ( x,y \right )}{\delta y}$

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. As the partial derivatives of a function $f(x, y)$ are themselves functions of ‘x’ and ‘y’, they in turn may have partial derivatives.

### 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. We can define the second order partial derivatives 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

$\frac{\partial f}{\partial x}$ = 0 and $\frac{\partial f}{\partial y}$ = 0 ..................(a)

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

### Solved Example

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

** Solution: **
Just like ordinary derivatives, partial derivatives follows some rule like product rule, quotient rule, chain rule etc.

### Partial Derivative Properties

If we have u = f(x,y) then, partial derivatives follows some rules as the ordinary derivatives.

**Product Rule:**

If u = f(x,y).g(x,y), then

$u_{x}$ = $\frac{\partial u}{\partial x}$ = $g\left ( x,y \right )$$\frac{\partial f}{\partial x}$$ + f\left ( x,y \right )$$\frac{\partial g}{\partial x}$

And, $u_{y}$ = $\frac{\partial u}{\partial y}$ = $g\left ( x,y \right )$$\frac{\partial f}{\partial y}$$ + f\left ( x,y \right )$$\frac{\partial g}{\partial y}$

**Quotient Rule:**

If u = $\frac{f(x,y)}{g(x,y)}$, where g(x,y) $\neq$ 0 then,

$u_{x}$ = $\frac{g\left ( x,y \right )\frac{\partial f}{\partial x}-f\left ( x,y \right )\frac{\partial g}{\partial x}}{\left [ g\left ( x,y \right ) \right ]^{2}}$

And, $u_{y}$ = $\frac{g\left ( x,y \right )\frac{\partial f}{\partial y}-f\left ( x,y \right )\frac{\partial g}{\partial y}}{\left [ g\left ( x,y \right ) \right ]^{2}}$

**Power Rule:**

If u = [f(x,y)]^{2} then, partial derivative of u with respect to x and y defined as

$u_{x} = n\left [ f\left ( x,y \right ) \right ]^{n - 1} $$\frac{\partial f}{\partial x}$

And, $u_{y} = n\left [ f\left ( x,y \right ) \right ]^{n - 1} $$\frac{\partial f}{\partial y}$

**Chain Rule:**

If u = f(x,y) is a function where, x = (s,t) and y = (s,t) then by the chain rule, we can find the partial derivatives u_{s} and u_{t} as:

$u_{s}$ = $\frac{\partial u}{\partial x}.\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial s}$

and $u_{t}$ = $\frac{\partial u}{\partial x}.\frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial t}$### Solved Examples

**Question 1: **f(x,y) = x^{2}y + sin x + cos y

** Solution: **

**Question 2: **xy + x^{2}

** Solution: **
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

$\frac{\partial f}{\partial x}$ = $\lim_{\delta x \rightarrow 0}$$\frac{f\left ( x + \delta x , y \right ) - f\left ( x,y \right )}{\delta x}$ is the first partial derivative of f with respect to x.

Then, $f_{xy}$ = $\frac{\partial }{\partial y}\left ( \frac{\partial f}{\partial x} \right )$ = $\frac{\partial ^{2}f}{\partial y \partial x }$ and $f_{yx}$ = $\frac{\partial }{\partial x}\left ( \frac{\partial f}{\partial y} \right )$ = $\frac{\partial ^{2}f}{\partial x \partial y }$

The terms $f_{xy}$ and $f_{yx}$ are called the mixed or cross partial derivative of f.### Solved Example

**Question: **Find the partial derivatives f_{xy} and f_{yx} of the function f(x,y) = $x^{3}y^{4} – y\ \sin(x)$

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

### Solved Examples

**Question 1: **If f(x,y) = xy and y = e^{x}, find $\frac{df}{dx}$.

** Solution: **
**Question 2: **If f(x,y) = 2x + 3y, where x = t and y = t^{2}, find $\frac{df}{dt}$.

** Solution: **

In mathematics, a

**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 $\frac{\partial f}{\partial x}$ keeping y terms as constant. Note that, it is not dx, instead it is $\partial x$.

$\frac{\partial f}{\partial x}$ is also known as f_{x}.

The partial derivative of a function f with respect to the variable x is variously denoted by $f'_x$, $f_x$, $\partial_x f$ or $\frac{\partial f}{\partial x}$

Here, $\partial$ is the symbol of partial derivative.

To find the partial derivative of function express in to form of "In", we follow the same procedure as finding the derivative of the function. But, this time when we calculate the partial derivative of the function with respect to one independent variable taking other as constant and follow the same thing with others.

Given below are some of the examples on Partial Derivatives.

Given function is f(x, y) = tan(xy) + sin x

Derivative of a function with respect to x is given as follows:

f_{x } = $\frac{\partial f}{\partial x}$ = $\frac{\partial}{\partial x}$$[\tan (xy) + \sin x]$

= $\frac{\partial}{\partial x}$$ [\tan(xy)] + $$\frac{\partial}{\partial x}$ $[\sin x]$

= sec^{2} (xy). y + cos x

= y sec^{2}(xy) + cos x

Now, Derivative of a function with respect to y. So, x is constant

f_{y} = $\frac{\partial f}{\partial y}$ = $\frac{\partial}{\partial y}$$ [\tan (xy) + \sin x]$

= $\frac{\partial}{\partial y}$$ [\tan (xy)] + $$\frac{\partial}{\partial y}$$[\sin x]$

= sec^{2} (xy) x + 0

= x sec^{2} (xy)

**Answer**: f_{x }= y sec^{2}(xy) + cos x and f_{y} = x sec^{2} (xy)

Given, $f = x^{2} (y – z) + y^{2}(z – x) + z^{2}(x – y)$

To find $\frac {\partial f} {\partial x}$ ‘y and z’ are held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’

$\frac {\partial f} {\partial x}$ = $2x(y - z) + 0(0 - 1) + 0(1 - 0)$

$\frac {\partial f} {\partial x}$ = $2xy - 2xz + 0 + 0$

$\frac {\partial f} {\partial x}$ = $2xy - 2xz$--------(i)

To find $\frac {\partial f} {\partial y}$ 'x and z’ is held constant and the resulting function of ‘y’ is differentiated with respect to ‘y’.

$\frac {\partial f} {\partial y}$ = $0(1 - z) + 2y(z - x) + 0(0 - 1)$

$\frac {\partial f} {\partial y}$ = $0 + 2yz - 2yx + 0$

$\frac {\partial f} {\partial y}$ = $2yz - 2yx$ -------(ii)

To find $\frac {\partial f} {\partial z}$ 'x and y’ is held constant and the resulting function of ‘z’ is differentiated with respect to ‘z’.

$\frac {\partial f} {\partial z}$ = $0(0 - 1) + 0(1 - 1) + 2z(x - y)$

$\frac {\partial f} {\partial z}$ = $0 + 0 + 2zx - 2zy$

$\frac {\partial f} {\partial z}$ = $2zx - 2zy$ ---------(iii)

Adding (i) , (ii) and (iii) we get

$\frac {\partial f} {\partial x}$ + $\frac {\partial f} {\partial y}$ + $\frac {\partial f} {\partial z}$ = $2xy - 2xz + 2yz - 2yx - 2zx - 2zy$

$\frac {\partial f} {\partial x}$ + $\frac {\partial f} {\partial y}$ + $\frac {\partial f} {\partial z}$ = $0$

Hence proved.

To find $\frac {\partial f} {\partial x}$ ‘y and z’ are held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’

$\frac {\partial f} {\partial x}$ = $2x(y - z) + 0(0 - 1) + 0(1 - 0)$

$\frac {\partial f} {\partial x}$ = $2xy - 2xz + 0 + 0$

$\frac {\partial f} {\partial x}$ = $2xy - 2xz$--------(i)

To find $\frac {\partial f} {\partial y}$ 'x and z’ is held constant and the resulting function of ‘y’ is differentiated with respect to ‘y’.

$\frac {\partial f} {\partial y}$ = $0(1 - z) + 2y(z - x) + 0(0 - 1)$

$\frac {\partial f} {\partial y}$ = $0 + 2yz - 2yx + 0$

$\frac {\partial f} {\partial y}$ = $2yz - 2yx$ -------(ii)

To find $\frac {\partial f} {\partial z}$ 'x and y’ is held constant and the resulting function of ‘z’ is differentiated with respect to ‘z’.

$\frac {\partial f} {\partial z}$ = $0(0 - 1) + 0(1 - 1) + 2z(x - y)$

$\frac {\partial f} {\partial z}$ = $0 + 0 + 2zx - 2zy$

$\frac {\partial f} {\partial z}$ = $2zx - 2zy$ ---------(iii)

Adding (i) , (ii) and (iii) we get

$\frac {\partial f} {\partial x}$ + $\frac {\partial f} {\partial y}$ + $\frac {\partial f} {\partial z}$ = $2xy - 2xz + 2yz - 2yx - 2zx - 2zy$

$\frac {\partial f} {\partial x}$ + $\frac {\partial f} {\partial y}$ + $\frac {\partial f} {\partial z}$ = $0$

Hence proved.

$f_{x}$ = $\frac{\partial f}{\partial x}$ = $\lim_{h\rightarrow 0}$$\frac{f(x + h,y) - f(x,y)}{h}$

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

$f_{y}$ = $\frac{\partial f}{\partial y}$ = $\lim_{h\rightarrow 0}$$\frac{f(x,y + h) - f(x,y)}{h}$

If u = f(x,y) is then the partial derivative of f with respect to x defined as $\frac{\partial f}{\partial x}$ and denoted by

$\frac{\partial f}{\partial x}$ = $\lim_{\delta x \rightarrow 0}$$\frac{f\left ( x + \delta x , y \right )-f\left ( x,y \right )}{\delta x}$

And, partial derivative of f with respect to y defined as $\frac{\partial f}{\partial y}$ and denoted by

$\frac{\partial f}{\partial y}$ = $\lim_{\delta y\rightarrow 0}$$\frac{f\left ( x,y+\delta y \right )-f\left ( x,y \right )}{\delta y}$

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. As the partial derivatives of a function $f(x, y)$ are themselves functions of ‘x’ and ‘y’, they in turn may have partial derivatives.

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. We can define the second order partial derivatives as follows

- $\frac {\partial } {\partial x} (\frac {\partial f} {\partial x})$ is denoted by $\frac {\partial ^{2} f} {\partial ^{2} x}$ or $f_{xx}$ or $f_{x}^{2}$
- $\frac {\partial } {\partial y} \left ( \frac {\partial f} {\partial y} \right )$ is denoted by $\frac {\partial ^{2} f} {\partial y^{2} x}$ or $f_{yy}$ or $f_{y}^{2}$
- $\frac {\partial } {\partial x} \left ( \frac {\partial f} {\partial y} \right )$ is denoted by $\frac {\partial ^{2} f} {\partial x. \partial y}$ or $f_{xy}$
- $\frac {\partial } {\partial y} \left ( \frac {\partial f} {\partial x} \right )$ is denoted by $\frac {\partial ^{2} f} {\partial y. \partial x}$ or $f_{yx}$

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.

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

$\frac{\partial f}{\partial x}$ = 0 and $\frac{\partial f}{\partial y}$ = 0 ..................(a)

If (x

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

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

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

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}

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

To find f_{y} , ‘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}

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

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

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

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

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

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

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

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

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

(i) f

f

To find f

f

To find f

f

(ii) f

f

To find f

f

To find f

f

(iii) f

f

To find f

f

To find f

f

(iv) f

f

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

f

To find f

f

If u = f(x,y).g(x,y), then

$u_{x}$ = $\frac{\partial u}{\partial x}$ = $g\left ( x,y \right )$$\frac{\partial f}{\partial x}$$ + f\left ( x,y \right )$$\frac{\partial g}{\partial x}$

And, $u_{y}$ = $\frac{\partial u}{\partial y}$ = $g\left ( x,y \right )$$\frac{\partial f}{\partial y}$$ + f\left ( x,y \right )$$\frac{\partial g}{\partial y}$

If u = $\frac{f(x,y)}{g(x,y)}$, where g(x,y) $\neq$ 0 then,

$u_{x}$ = $\frac{g\left ( x,y \right )\frac{\partial f}{\partial x}-f\left ( x,y \right )\frac{\partial g}{\partial x}}{\left [ g\left ( x,y \right ) \right ]^{2}}$

And, $u_{y}$ = $\frac{g\left ( x,y \right )\frac{\partial f}{\partial y}-f\left ( x,y \right )\frac{\partial g}{\partial y}}{\left [ g\left ( x,y \right ) \right ]^{2}}$

If u = [f(x,y)]

$u_{x} = n\left [ f\left ( x,y \right ) \right ]^{n - 1} $$\frac{\partial f}{\partial x}$

And, $u_{y} = n\left [ f\left ( x,y \right ) \right ]^{n - 1} $$\frac{\partial f}{\partial y}$

If u = f(x,y) is a function where, x = (s,t) and y = (s,t) then by the chain rule, we can find the partial derivatives u

$u_{s}$ = $\frac{\partial u}{\partial x}.\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial s}$

and $u_{t}$ = $\frac{\partial u}{\partial x}.\frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial t}$

Now, finding out f_{x }first keeping y as constant

f_{x }= $\frac{\partial f}{\partial x}$ = (2x) y + cos x + 0

= 2xy + cos x

when we keep y as constant cos y becomes a constant so its derivative becomes zero.

Similarly, finding f_{y}

f_{y }= $\frac{\partial f}{\partial y}$ = x^{2} + 0 + (-sin y)

= x^{2} - sin y

Lets find out the partial derivative with respect to x keeping y as constant.

$\frac{\partial f}{\partial x}$ = y + 2x

Here,
while differentiating the xy term, we treat y as a constant. So, the
differentiation of x is 1 which will simply get multiplied by y and the
second term is x^{2} in terms of x only so its derivative is simply 2x.

Now, for the same problem we try to find out partial derivative with respect to y or f_{y}

f_{y }= $\frac{\partial f}{\partial y}$ = x + 0

treating x term as constant the second term x^{2} becomes a constant so its derivative with respect to y is 0.

f_{y }= x

$\frac{\partial f}{\partial x}$ = $\lim_{\delta x \rightarrow 0}$$\frac{f\left ( x + \delta x , y \right ) - f\left ( x,y \right )}{\delta x}$ is the first partial derivative of f with respect to x.

Then, $f_{xy}$ = $\frac{\partial }{\partial y}\left ( \frac{\partial f}{\partial x} \right )$ = $\frac{\partial ^{2}f}{\partial y \partial x }$ and $f_{yx}$ = $\frac{\partial }{\partial x}\left ( \frac{\partial f}{\partial y} \right )$ = $\frac{\partial ^{2}f}{\partial x \partial y }$

The terms $f_{xy}$ and $f_{yx}$ are called the mixed or cross partial derivative of f.

To find f_{xy} and f_{yx} first we have to find f_{x} and f_{y}. So, 'y' is held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’

$\frac {\partial f} {\partial x}$ = $y^{4}(3x^{2}) - y.\ (\cos x)$

$\frac {\partial f} {\partial x}$ = $3x^{2} y^{4} - y. (\cos x)$

To find $\frac {\partial f} {\partial y}$,

$\frac {\partial f} {\partial y}$ = $x^{3}(4y^{3}) - \sin (x)$

$\frac {\partial f} {\partial y}$ = $4x^{3}y^{3} - \sin (x)$

Now, f_{xy} = $\frac{\partial}{\partial y} [\frac {\partial f} {\partial y}]$

= $\frac{\partial}{\partial y}$ [ 3x^{2}y^{4} - y cos x]

= 12x^{2}y^{3} - cos x

Again, f_{yx} = $\frac{\partial}{\partial x} [\frac {\partial f} {\partial x}]$

= $\frac{\partial}{\partial x}$ [4x^{3}y^{3} - sin x]

= 12x^{2}y^{3} - cos x

$\frac {\partial f} {\partial x}$ = $y^{4}(3x^{2}) - y.\ (\cos x)$

$\frac {\partial f} {\partial x}$ = $3x^{2} y^{4} - y. (\cos x)$

To find $\frac {\partial f} {\partial y}$,

$\frac {\partial f} {\partial y}$ = $x^{3}(4y^{3}) - \sin (x)$

$\frac {\partial f} {\partial y}$ = $4x^{3}y^{3} - \sin (x)$

Now, f

= $\frac{\partial}{\partial y}$ [ 3x

= 12x

Again, f

= $\frac{\partial}{\partial x}$ [4x

= 12x

- 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 to total differentiation. Then the total partial derivative of u with respect to t is $\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$
- If f is a function defined as f(x), where x(u,v) then $\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u}$ and $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v}$
- Suppose f = f(x,y) and y is a implicit function it means y is itself a function of x, then $\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}$
- If f(x,y) where x(u,v) and y(u,v), then $\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}$ and $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}$

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

$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}$

Now, $\frac{\partial f}{\partial x}$ = y, $\frac{\partial f}{\partial y}$ = x and $\frac{dy}{dx}$ = e^{x}

Then, $\frac{df}{dx}$ = y + x e^{x} = e^{x} + x e^{x} = e^{x} (1 + x)

$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}$

Now, $\frac{\partial f}{\partial x}$ = y, $\frac{\partial f}{\partial y}$ = x and $\frac{dy}{dx}$ = e

Then, $\frac{df}{dx}$ = y + x e

We know that $\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$

Then, $\frac{\partial f}{\partial x}$ = 2, $\frac{\partial f}{\partial y}$ = 3, $\frac{dx}{dt}$ = 1, $\frac{dy}{dt}$ = 2t

So, we get

$\frac{df}{dt}$ = 2.1 + 3.(2t) = 2 + 3t

Then, $\frac{\partial f}{\partial x}$ = 2, $\frac{\partial f}{\partial y}$ = 3, $\frac{dx}{dt}$ = 1, $\frac{dy}{dt}$ = 2t

So, we get

$\frac{df}{dt}$ = 2.1 + 3.(2t) = 2 + 3t

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