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Total Derivative

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  • Difficulty Level : Basic
  • Last Updated : 19 Jul, 2022
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The total derivative of a function f at a point is approximation near the point of function w.r.t. (with respect to) its arguments (variables). Total derivative never approximates the function with a single variable if two or more variables are present in the function. Sometimes, the Total derivative is the same as the partial derivative or ordinary derivative of the function.

For Composite function:

In general composite, the function is nothing but a function of two or more dependent variables which depend upon any common variable t. Composite function values are obtained from both variables. 

If u= f(x,y), where x and y are dependent variables at t, then we can also express u as a function of t. By substituting the value of x, y in f(x,y). Thus, we find the ordinary derivative which is called the total derivative of u

Now, to find \frac{du}{dt}    without actually substituting the value of x and y in f(x,y). 

 \frac{du}{dt} =\frac{\partial u }{\partial x}. \frac{\partial x }{\partial t} + \frac{\partial u }{\partial y}. \frac{\partial y }{\partial t}

Similarly, If u = f(x,y,z) where x, y, z are all function of a variable t , then chain rule is:

 \frac{du}{dt} =\frac{\partial u }{\partial x}. \frac{\partial x }{\partial t} + \frac{\partial u }{\partial y}. \frac{\partial y }{\partial t} + \frac{\partial u }{\partial z}. \frac{\partial z }{\partial t}

Question:  Given,  u =  sin\frac{x}{y} , x = e^{t}, y= t^{2},  find  \frac{du}{dt}           as a function of t.  Verify your result by direct substitution .

Solution: We have,  \frac{du}{dt} = \frac{\partial u }{\partial x}. \frac{\partial x }{\partial t} + \frac{\partial u }{\partial y}. \frac{\partial y }{\partial t}

                                            = cos\frac{x}{y} .  \frac{1}{y} . _e{t} + (cos\frac{x}{y}) (\frac{-x}{_y{2}^{ }}) .2t

                                               putting values of x and y in above equations

                                            =cos\frac{_e{t}}{_t{2}} .  \frac{_e{t}}{_t{2}} -2cos\frac{_e{t}}{_t{2}} . _e{t}._t{3}

                        \frac{du}{dt} = (t-\frac{2}{_t{3}})._e{t}.cos\frac{_e{t}}{_t{2}}           

Question: Given, f(x,y) =exsiny , x=t3+1 and y=t4+1. Then df/dt at t =1. 

Solution: Let f(x,y) =exsiny 

 \frac{df}{dt} =\frac{\partial u }{\partial x}. \frac{\partial x }{\partial t} + \frac{\partial u }{\partial y}. \frac{\partial y }{\partial t}

                     = exsiny.(3t2) + cosy .ex .(4t3

 As we know , x= t3+1 and y= t4+1

x and y values at t =1, x=2 and y=2

             \frac{df}{dt} = (e^{2})(sin2)(12) + (cos2)(e^{2})(32)

                                =(2.718)2(0.0349)(12) +(0.9994)(2.718)2(32)

                                = 238.97

For Implicit function:

The implicit function is a function whose variables are not completely independent variables. Let a function f(x,y) where x is independent variable but y is x dependent variable. 

If f(x, y)= c ( constant )be an implicit function and relation between x and y exists which defines as a differentiable function of x

Here, f(x,y) =constant 

For implicit function let us consider, x an independent variable, and y is a function of x

          f(x,y) = c           ……..eq (1)

by definition of total differential coefficient.

Question: If u = xlogxy where x3 +y3+3xy=1, find du/dx.

Solution: We have x3 +y3+3xy=1    ……….(1)

\frac{du}{dx}=\frac{\partial u}{\partial x}.\frac{dy}{dx} +\frac{\partial u}{\partial y}.\frac{dy}{dx}

          = (logxy +1) + \frac{x}{y}.\frac{dy}{dx} ..........(2)         

from eq……….(1)

\frac {dy}{dx} = -\frac{\frac{∂f}{∂x}} { \frac{∂f}{∂y}}

\frac{dy}{dx} = -\frac {(3x^{2} + 3y)}{(3y^{2} + 3x)}

= -\frac{(x^{2} +y)}{(y^{2} +x)}t

after putting value in eq (2)

\frac{du}{dx} = (logxy +1) - (\frac{x}{y})\frac {(x^{2}+y)}{(y^{2}+x)} .

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