Total Derivative (original) (raw)

Last Updated : 15 Sep, 2025

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, . . . , xn: f = f(x1, x2, . . . , xn)

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 xi​.
• \frac{dx_i}{dt} is the derivative of​ xi 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 thetotal derivativeof 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:

**Related Articles:

• **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) = 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

**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 + ey, 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·ey, 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.