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Partial derivative is generally used in vector analysis and differential geometry. There are different functions given in the question with some of the functions depending on two or more variables in vector analysis and differential geometry. If the function depends on only one variable then it is known as ordinary differentiation. If the function depends upon more than one variable then a derivative taking either with respect to two variables is called partial differentiation.
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Key Takeaways - Partial derivative, Quotient rule, power rule, product rule, chain rule, partial derivative, dependent variable
Also read: Isosceles Triangle Theorems
Partial Derivative Definition
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The function f and its variable x and y, the representation will be f(x,y) i.e. the function f depends on x and y. The variables x and y are independent of each other. If we differentiate the function with respect to x then y is constant and if we differentiate with respect to y then x is constant.
Partial Derivative Symbol
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The symbol of partial differentiation is ∂ i.e. called dell. The f(x,y) the partial differentiation concerning with x is ∂f/∂x, then y is keep constant. The ∂x and dx are not same.
The function f depends on both x and y. The differentiate f with respect to x partially and keep y is constant by using limit function.
Also read: Difference between Sequence and Series
Partial Derivation Formula
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fx= \(\frac{\partial f}{\partial x}\) = (f (x+hy) – f(xy)) / h
The differentiate f with respect to y partially and keep x is constant by using limit function.
fy= \(\frac{\partial f}{\partial y}\) = (f (xy+h) – f(xy)) / h
Also Read: Differential equation
Partial Differentiation
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Partial differentiation is the process of determining a function's partial derivatives. When we examine one of the tangent lines of the graph of a given function and find its slope, we will use partial differentiation.
Partial Derivatives Rules
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Product Rule
If u= f(x,y) . g(x,y) then product rule will be
ux = \(\frac{du}{dx}\) = g(x,y)\(\frac{df}{dx}\) + f(x,y)\(\frac{dg}{dx}\)
uy = \(\frac{du}{dy}\) = g(x,y)\(\frac{df}{dy}\) + f(x,y)\(\frac{dg}{dy}\)
Quotient Rule
Suppose, u= f(x,y)/ g(x,y), where g(x,y) ? 0
ux = \(\frac{g (x,y) \frac{df}{dx} - f (x,y) \frac{dg}{dx}}{ [g(xy)]^2} \)
uy = \(\frac{g (x,y) \frac{df}{dy} - f (x,y) \frac{dg}{dy}}{ [g(xy)]^2} \)
Power Rule
If f = (x,y)n , then partial derivatives will be
ux = n|f(x,y)|n-1 ∂f/∂x
uy = n|f(x,y)|n-1 ∂f/∂y
Chain Rule for the Independent Variable
If x=g(t) and y=h(t) are differentiable functions of t, and z = f(x, y) is a differentiable function of both x and y. Then, z can be written as z = f(g(t), h(t)) - a differentiable function of t.
The partial derivative of the function with respect to the variable t will be given as follows:
∂z/∂t = ∂z/∂x × ∂x/∂t + ∂z/∂y × ∂y/∂t
Chain Rule for The Dependent Variable
Assume, x = g (u, v) and y = h (u, v) are the differentiable functions of the given variables u and v,
Similarly, z = f (x, y) is a differentiable function of x and y. Thus z can be further defined as z = f (g (u, v), h (u, v)), which is a differentiable function of both u and v.
With respect to the provided variables, the partial derivative of the above functions will be written as:
∂z/∂u = ∂z/∂x × ∂x/∂u + ∂z/∂y × ∂y/∂u
∂z/∂v = ∂z/∂x × ∂x/∂v + ∂z/∂y × ∂y/∂v
The Partial Derivative of Natural Logarithm (In)
The approach for calculating the partial derivative of natural logarithm "In" is the same as for calculating the derivative of any normal function. The partial derivative of the function is calculated with respect to one independent variable, and the others are taken as constant.
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Things to Remember
- The symbol of partial differentiation is ∂ i.e. called dell.
- The f(x,y) the partial differentiation concerning with x is ∂f/∂x, then y is keep constant.
- The ∂x and dx are not same.
- The function f depends on both x and y. The differentiate f with respect to x partially and keep y is constant by using limit function.
fx = \(\frac{\partial f}{\partial x}\) = (f (x+hy) – f(xy)) / h
- With respect to the given variables, the partial derivative of the above functions will be written as:
∂z/∂u = ∂z/∂x × ∂x/∂u + ∂z/∂y × ∂y/∂u
∂z/∂v = ∂z/∂x × ∂x/∂v + ∂z/∂y × ∂y/∂v
Also read: First Order Differential Equation
Solved Questions
Ques. Find the first and second partial derivatives of z = x3 + y3 – 3axy. (4 marks)
Ans: Given- z = x3 + y3 – 3axy -------- (1)
Partially differentiating eq (1) w.r.t. x
\(\frac{\partial z}{\partial x}\) = 3x2 – 3ay ----------------- (2)
Again partially differentiate eq (2) w.r.t. x
∂2z /∂x2 = 6x --------------------- (3)
Again partially differentiate eq (2) w.r. t. y
∂2z/\(\frac{\partial y}{\partial x}\) = -3a
Partially differentiating eq (1) w.r.t. y
\(\frac{\partial z}{\partial y}\) = 3y2 – 3ax ---------------- (4)
Again partially differentiate eq (4) w.r.t. y
∂2z /∂y2 = 6y
Again partially differentiate eq (4) w.r. t. x
∂2z/\(\frac{\partial x}{\partial y}\) = -3a
We observed that,
∂2z/\(\frac{\partial y}{\partial x}\) = ∂2z/\(\frac{\partial x}{\partial y}\)
Ques. If u = x2y + y2z + z2x , then prove that \(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\) = (x+y+z)2 (4 marks)
Ans: Given: u = x2y + y2z + z2x ------------------ (1)
Partially differentiating eq (1) by x keeping y and z are constants
\(\frac{\partial u}{\partial x}\) = 2xy + 0 + z2.1
\(\frac{\partial u}{\partial x}\) = 2xy + z2 ------------------------- (A)
Similarly, differentiate eq (1) partially by y and keeping x and z constant
\(\frac{\partial u}{\partial y}\) = x2 + 2yz --------------------------- (B)
Similarly, differentiate eq (1) partially by z and keeping x and y constant
\(\frac{\partial u}{\partial z}\) = y2 + 2zx ---------------------------- (C)
Add (A) + (B) + (C)
\(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\) = 2xy + z2 + x2 + 2yz + y2 + 2zx
\(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\) = + x2 + y2 + z2+ 2xy + 2yz + 2zx
Hence,
\(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\) = (x + y + z)2
Ques. u = log (x3 + y3 + z3 – 3xyz) then show that \(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\) = \(\frac{3}{x+y+z}\) (\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + \(\frac{\partial}{\partial z}\) )2 u = -9/ (x+y+z)2 (4 marks)
Ans: u = log (x3 + y3 + z3 – 3xyz) ----------------------- (1)
Partially differentiate eq (1) w.r.t. x by keeping y and z constant
\(\frac{\partial u}{\partial x}\) = 1x3 + y3 + z3 – 3xyz . 3x2 – 3yz -------------------------- ddx logx = 1/x
\(\frac{\partial u}{\partial x}\) = (3x2-3yz) / (x3 + y3 + z3 – 3xyz) ------------------------- ddx xn = nxn-1 ---------- (A)
Similarly,
\(\frac{\partial u}{\partial y}\) = (3y2 -3xz) / (x3 + y3 + z3 – 3xyz) ------------------------- (B)
\(\frac{\partial u}{\partial z}\) = (3z2 -3xy) / (x3 + y3 + z3 – 3xyz) -------------------------- (C)
Add (A), (B) and (C)
We get,
\(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\) = (3x2-3yz + 3y2 -3xz + 3z2 -3xy) / (x3 + y3 + z3 – 3xyz)
Taking 3 common from numerator
\(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\) = 3 (x2 + y2+ z2-xy -yz –xz) / (x3 + y3 + z3 – 3xyz)
There is one standard formula,
x3 + y3 + z3 – 3xyz = (x+y+z) (x2 + y2+ z2-xy -yz –xz)
\(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\) = \(\frac{3}{x+y+z}\) ----------------------- (D)
(\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + ∂z )2 u = (\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + \(\frac{\partial}{\partial z}\)) (\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + \(\frac{\partial}{\partial z}\)) u
(\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + ∂z )2 u = (\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + \(\frac{\partial}{\partial z}\)) (\(\frac{\partial u}{\partial x}\) + \(\frac{\partial u}{\partial y}\) + \(\frac{\partial u}{\partial z}\))
(\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + ∂z )2 u = (\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + \(\frac{\partial}{\partial z}\)) \(\frac{3}{x+y+z}\) ---------- [from (3)]
(\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + ∂z )2 u = (\(\frac{\partial}{\partial x}\) \(\frac{3}{x+y+z}\) + \(\frac{\partial}{\partial y}\) \(\frac{3}{x+y+z}\) + \(\frac{\partial}{\partial z}\) \(\frac{3}{x+y+z}\) )
\(\frac{\partial}{\partial x}\) 1/x = -1/x2
(\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + \(\frac{\partial}{\partial z}\) )2 u = -3/(x+y+z)2 - 3/(x+y+z)2 - 3/(x+y+z)2
(\(\frac{\partial}{\partial x}\) + \(\frac{\partial}{\partial y}\) + \(\frac{\partial}{\partial z}\) )2 u = -9/ (x + y + z)2
Ques. If z = f (x + ct) + Φ (x - ct), prove that ∂2z/ ∂t2 = c2 . ∂2z/ ∂x2 (4 marks)
Ans: z = f (x + ct) + Φ (x - ct) ------------------------ (1)
Partially differentiating eq (1) w.r.t. x,
\(\frac{\partial z}{\partial x}\) = f ’ (x + ct) + Φ’ (x - ct) --------------------(2)
Again, partially w.r.t. x
∂2z/ ∂x2 = f ’’ (x + ct) + Φ’’ (x - ct) ------------------- (3)
Partially differentiating eq (1) w.r.t. t
\(\frac{\partial z}{\partial t}\) = c f ’ (x + ct) + Φ’ (x - ct) (-c)
\(\frac{\partial z}{\partial t}\) = c f ’ (x + ct) -c Φ’ (x - ct) ------------------------ (4)
Again, partially w.r.t. t
∂2z/ ∂t2 = c2 f ’’ (x + ct) - c2 Φ’’ (x - ct)
∂2z/ ∂t2 = c2 [f ’’ (x + ct) - Φ’’ (x - ct)]
∂2z/ ∂t2 = c2 . ∂2z/ ∂x2 -------------- [from(3)]
Ques. What is the difference between differentiation and partial differentiation prove with example (4 marks)
Ans: Consider equation,
u= xy2 + xy – x2 ------------------------------ (1)
Differentiate w.r.t. x
du/dx = x.2y .(dy/dx) + y2 .(1) + x. (dy/dx) + y(1) - 2x
du/dx = 2xy (dy/dx) + y2 + x(du/dx) + y – 2x
Now differentiate partially w.r.t. x
∂u/∂x = y2 (1) + y(1) – 2x
∂u/∂x = y2 + y – 2x
In partial differentiation, if we differentiate w.r.t. x then we keep other variables as a constant
Ques. u = log (tanx + siny + cosz) Find ∂u/∂x (4 marks)
Ans: u = log (tanx + siny +cosz)
Differentiating ‘u’ partially w.r.t x
∂u/∂x = \(\frac{1}{tanx+siny+cosz}\) ∂x (tanx + siny +cosz)
∂u/∂x = \(\frac{1}{tanx+siny+cosz}\) sec2x
∂u/∂x = sec2x / (tanx + siny +cosz)
Ques. u = log (tanx + siny +cosz) Find ∂u/∂y (4 marks)
Ans: u = log (tanx + siny +cosz)
Differentiating ‘u’ partially w.r.t y
∂u/∂y = \(\frac{1}{tanx+siny+cosz}\) ∂y (tanx + siny +cosz)
∂u/∂y = \(\frac{1}{tanx+siny+cosz}\) cos y
∂u/∂y = cosy / (tanx + siny +cosz)
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