Total Derivative

Total Derivative of a function measures how that function changes as all of its input variables change. For function f at a point is an approximation near the point of the function w.r.t. (with respect to) its arguments (variables).

It is an approximation of the actual change in the function and is used to determine the total change in the dependent variable when all the independent variables are varied.

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 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, 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}

Difference Between Total Derivative and Partial Derivative

Common difference between total and partial derivative are:

Read More,

• Differentiation in Calculus
• Derivatives
• Partial Differential Equation
• Derivative Formulas
• Derivative Rules

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 = 𝑥^2 𝑦^3 , 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 = 2xy ^3 , ∂u / ∂y = 3x ^2 y ^2

differentiate x and y with respect to t:

dx/dt = −sin(t) , dy/dt = cos(t)

substitute into the total derivative formula:

du/dt = 2xy ^3 (−sin(t)) + 3x ^2 y ^2 (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=x ^3+y ^2 , where x=ln(t) and y=t^2 , find du/dt

Solution: ​

We have,

u=x ^3+y ^2

x=ln(t) , y=t^2

total derivatives:

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

Partial derivatives:

∂u /∂x= 3x ^2 , ∂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 +4t ^3