Total Derivative

The Total Derivative of a function measures how that function changes as all of its input variables change.

• For a multivariable function f, the total derivative at a point provides a linear approximation of how f varies with respect to its arguments near that point.
• It approximates the actual change in the dependent variable when all the independent variables are varied simultaneously.
• In essence, it tells us how small changes in each variable combine to affect the value of the function.

Formula for Total Derivative

Suppose f is a function of n variables, x₁, x₂, . . . , x_n: f = f(x₁, x₂, . . . , x_n)

The total derivative of f with respect to a variable t (which could represent time or another parameter) is given by:

\frac{df}{dt} = \sum_{i=1}^{n} \frac{\partial f}{\partial x_i} \frac{dx_i}{dt}

Here:

• \frac{\partial f}{\partial x_i}​ is the partial derivative of f with respect to x_i​.
• \frac{dx_i}{dt} is the derivative of​ x_i with respect to t.

Total Derivative of Composite Function

In general composite, the function is nothing but a function of two or more dependent variables that depend upon a 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, and z are all functions of a variable t, then the 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}

Total Derivative vs Partial Derivative

The common difference between total and partial derivatives are:

Total Derivative	Partial Derivative
Measures the rate of change of a function with respect to one variable while considering the effect of all other variables changing as well.	Measures the rate of change of a function with respect to one variable while keeping other variables constant.
d/dt​ or Df	∂/∂x or ∂f/∂x
Considers the dependencies of all other variables.	Considers one variable at a time, treating others as constants.
Used in contexts where variables are interdependent, such as in differential equations and dynamic systems.	Used in multivariable functions and fields like physics and economics, where one variable's effect is isolated.
Represents the slope of the tangent line to the curve of the function when all variables are allowed to change.	Represents the slope of the tangent line to the curve of the function in a specific direction, holding other variables fixed.

Solved Problems on Total Derivative

Question 1: 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 the 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 2: Given, f(x,y) = e^xsiny , x = t³+1 and y = t⁴+1. Then df/dt at t = 1. 

Solution:

Let f(x,y) = e^xsiny 

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

= e^xsiny.(3t²) + cosy .e^x .(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

Question 3: If u = x . log(xy) 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{dy}{dx} = -\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)} .

Question 4: If u = x²y³, where x = cos(t) and y = sin(t), find du/dt.

Solution:

We have, 𝑢 = 𝑥² 𝑦³ , x = cos(t) , y = sin(t)

The total derivative du/dt is

du/dt = ∂u /∂x ⋅ dx/dt + ∂u / ∂y ⋅ dy/dt

calculate the partial derivatives:

∂u /∂x = 2xy³ , ∂u / ∂y = 3x² y²

differentiate x and y with respect to t:

dx/dt = −sin(t) , dy/dt = cos(t)

substitute into the total derivative formula:

du/dt = 2xy³ (−sin(t)) + 3x² y² (cos(t) )

Substitute x = cos(t) and y = sin(t):

du/dt = 2(cos(t))(sin(t))³(−sin(t))+3(cos(t)) ² (sin(t))² (cos(t))

Simplify,

du/dt = −2cos(t)(sin(t)) ⁴ +3(cos(t)) ³ (sin(t))²

Question 5: If u = x³+ y² , where x = ln(t) and y = t² , find du/dt

Solution: ​

We have,

u = x ³+ y²

x = ln(t) , y = t²

total derivatives:

du/dt = ∂u /∂x ⋅ dx/dt + ∂u / ∂y ⋅ dy/dt

Partial derivatives:

∂u /∂x = 3x² , ∂u / ∂y = 2y

Derivatives of x and y:

dx/dt = 1/ t , dy/dt = 2t

simplify:

du/dt = 3(ln(t)) ² . 1/ t +2(t ² )⋅2t

= 3(ln(t)) ² / t +4t³

Unsolved Questions on Total Derivative

Question 1: Given u = x²y + e^y, x = t² + 1, y = sin(t). Find du/dt as a function of t.

Question 2: If f(x, y) = ln(x² + y²), x = cos(t), y = sin(t). Find df/dt at t = π/4.

Question 3: Let u = x·e^y, where x = t² – 1 and y = ln(t + 1). Find du/dt.

Question 4: If u = x² + y² + z²,where x = t, y = t², z = sin(t). find du/dt.

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