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🚀 Master Engineering Mathematics with Confidence!
📘 Step-by-step Engineering Mathematics solutions
❓ Doubt solving
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🎯 B.Tech | Diploma
Link - https://youtube.com/?si=Dtx7Vs6ibkpDX-vy

09/09/2026

433 | Inverse Jacobian Solved | If u=xyz, v=x²+y²+z², w=x+y+z Find ∂(x,y,z)/∂(u,v,w) | Engineering Maths

03/09/2026

432 | Jacobian Problem Solved | If y₁=cosx₁, y₂=sinx₁cosx₂, y₃=sinx₁sinx₂cosx₃ Prove J | Engineering
Learn Jacobian of Three Trigonometric Functions with complete step by step proof. This video is made for http://B.Tech 1st Year, BSc Maths, MSc, JEE Advanced, CUET PG, and GATE Engineering Mathematics students who need to master determinant based problems on Jacobian and spherical coordinate transformations.
00:00 If y₁ = cosx₁ , y₂=sinx₁cosx₂, y₃ = sinx₂sinx₁cosx₃ then show that the Jacobian of y₁, y₂,y₃ with respect to x₁,x₂,x₃ is -sin³x₁sin²x₂sinx₃
01:30 Jacobian of y₁, y₂,y₃ with respect to x₁,x₂,x₃
01:40 Trick to write Formula of Jacobian of y₁, y₂,y₃ with respect to x₁,x₂,x₃
03:00 Theorem of Jacobian
05:00 Jacobian of y₁, y₂,y₃ with respect to x₁,x₂,x₃ is -sin³x₁sin²x₂sinx₃

We are given y₁ = cosx₁, y₂ = sinx₁cosx₂, y₃ = sinx₁sinx₂cosx₃. We have to show that the Jacobian of y₁, y₂, y₃ with respect to x₁, x₂, x₃ is equal to minus sin³x₁ sin²x₂ sinx₃. This pattern appears in spherical coordinates and multiple integrals.

00:00 If y₁ = cosx₁ , y₂=sinx₁cosx₂, y₃ = sinx₂sinx₁cosx₃ then show that the Jacobian of y₁, y₂,y₃ with respect to x₁,x₂,x₃ is -sin³x₁sin²x₂sinx₃
This is the standard transformation from spherical to cartesian like coordinates. To prove it we need to find all 9 partial derivatives and evaluate the 3x3 determinant.

01:30 Jacobian of y₁, y₂,y₃ with respect to x₁,x₂,x₃
Formula:
∂(y₁,y₂,y₃)/∂(x₁,x₂,x₃) = determinant of 3x3 matrix
Row1: ∂y₁/∂x₁ ∂y₁/∂x₂ ∂y₁/∂x₃
Row2: ∂y₂/∂x₁ ∂y₂/∂x₂ ∂y₂/∂x₃
Row3: ∂y₃/∂x₁ ∂y₃/∂x₂ ∂y₃/∂x₃

Find partials:
∂y₁/∂x₁ = minus sinx₁, ∂y₁/∂x₂ = 0, ∂y₁/∂x₃ = 0
∂y₂/∂x₁ = cosx₁cosx₂, ∂y₂/∂x₂ = minus sinx₁sinx₂, ∂y₂/∂x₃ = 0
∂y₃/∂x₁ = cosx₁sinx₂cosx₃, ∂y₃/∂x₂ = sinx₁cosx₂cosx₃, ∂y₃/∂x₃ = minus sinx₁sinx₂sinx₃

01:40 Trick to write Fo

27/08/2026

431 | Jacobian Problem Solved | If y₁=x₂x₃/x₁, y₂=x₃x₁/x₂, y₃=x₁x₂/x₃ Prove J=4 | Engineering Maths

14/08/2026

430 | Jacobian Verification | If u=rcosΘ, v=rsinΘ Prove J₁.J₂=1 | Polar to Cartesian | Engineering Mathematics

09/08/2026

429 | Jacobian Full Concept + Solved Example | Definition, Formula, Properties and Problems
Partial DIFFERENTIATION playlist - https://youtube.com/playlist?list=PLT2xRiy9TL2gDydKo8lzx-dQLLm3IFG8A&si=JBJbDytucgcuOf_h

Notes- https://youtube.com/shorts/lfcZvlAEBO0?si=5Cmw3cAaaTpHLFU8
Learn Jacobian of Transformation with complete theory and one solved numerical. This video is made for http://B.Tech 1st Year, BSc Maths, MSc, JEE Advanced, CUET PG, and GATE Engineering Mathematics students who need clear concept and exam oriented problems of Jacobian.
00:00 Defination of Jacobian
01:30 Necessary condition for Jacobian
01:50 Formula of Jacobian
02:30 Jacobian of three functions
03:45 find the Jacobian of u,v with respect to x & y variable, where u=e^xsiny & v = x+log(siny)
04:20 Trick to write Formula of Jacobian of u,v with respect to x,y
06:00 Property of Jacobian

In this lecture we cover definition, necessary condition, formula for 2 and 3 functions, shortcut trick to write Jacobian, and key properties. At the end we solve a full example to find Jacobian of u and v with respect to x and y.

00:00 Definition of Jacobian
Jacobian is the determinant of the matrix of all first order partial derivatives of a set of functions with respect to a set of variables.
If u = f(x,y) and v = g(x,y) then Jacobian of u,v with respect to x,y is written as J or ∂(u,v)/∂(x,y).
It is used in change of variables, area, volume and in evaluation of multiple integrals.

01:30 Necessary condition for Jacobian
The number of functions must be equal to the number of variables for Jacobian to exist as a square determinant.
For ∂(u,v)/∂(x,y) we need 2 functions and 2 variables.
For ∂(u,v,w)/∂(x,y,z) we need 3 functions and 3 variables.

01:50 Formula of Jacobian
For two functions:
∂(u,v)/∂(x,y) = determinant of matrix
∂u/∂x ∂u/∂y
∂v/∂x ∂v/∂y
= ∂u/∂x times ∂v/∂y minus ∂u/∂y times ∂v/∂x

02:30 Jacobian of three functions
∂(u,v,w)/∂(x,y,z) = determinant of 3x3 matrix of partial derivatives
This is used in triple integrals and change of variables in 3D.

03:45 find the Jacobian of u,v with respect to x & y variabl

02/08/2026

428 | Implicit Differentiation Solved | If x^y + y^x = a^b Find dy/dx | Engineering Maths
Partial DIFFERENTIATION playlist - https://youtube.com/playlist?list=PLT2xRiy9TL2gDydKo8lzx-dQLLm3IFG8A&si=JBJbDytucgcuOf_h

Notes- https://youtube.com/shorts/lfcZvlAEBO0?si=5Cmw3cAaaTpHLFU8
Learn Implicit Differentiation and Logarithmic Differentiation with a complete step by step solution. This video is made for http://B.Tech 1st Year, BSc Maths, MSc, JEE Advanced, CUET PG, and GATE Engineering Mathematics students who need to solve problems where y is not explicitly defined in terms of x.
00:00 If x^y + y^x=a^b, find dy/dx
00:30 Implicit Equation
01:00 Logarithmic differentiation
03:00 dy/dx

If you are asked to find dy/dx for the equation x to the power y plus y to the power x equals a to the power b, this lecture explains how to handle both variable base and variable power using logarithmic differentiation. This type of question is very common in semester exams and competitive tests.

00:00 If x^y + y^x = a^b, find dy/dx
We are given an implicit equation where both x and y appear in base and exponent. Direct differentiation is not possible. So we differentiate both terms separately using logarithmic differentiation and then apply sum rule.

00:30 Implicit Equation
Differentiate both sides w.r.t x:
d/dx of x^y + d/dx of y^x = d/dx of a^b
Right side is constant so derivative is 0.
Let P = x^y and Q = y^x
So dP/dx + dQ/dx = 0

01:00 Logarithmic differentiation
For P = x^y:
Take log: ln P = y ln x
Differentiate: 1/P times dP/dx = dy/dx times ln x + y times 1/x
So dP/dx = x^y times dy/dx times ln x + y x^(y-1)

For Q = y^x:
Take log: ln Q = x ln y
Differentiate: 1/Q times dQ/dx = 1 times ln y + x times 1/y times dy/dx
So dQ/dx = y^x times ln y + x y^(x-1) times dy/dx

03:00 dy/dx
Now add dP/dx + dQ/dx = 0
x^y times dy/dx times ln x + y x^(y-1) + y^x times ln y + x y^(x-1) times dy/dx = 0

Group terms with dy/dx:
dy/dx times x^y ln x + x y^(x-1) + y x^(y-1) + y^x ln y = 0

Final Answer:
dy/dx = minus y x^(y-1) + y^x ln y over x^y ln x + x y^(x-1)

This is the required derivative. Write it in this form for full marks in exams.

What you

23/07/2026

427 | Related Rates Problem | Find dy/dt When 2xy-3x²y is Constant | Engineering Maths
Partial DIFFERENTIATION playlist - https://youtube.com/playlist?list=PLT2xRiy9TL2gDydKo8lzx-dQLLm3IFG8A&si=JBJbDytucgcuOf_h

Notes- https://youtube.com/shorts/09NmFTgSHh8?si=x50inz6L5miN4VAP
Learn Related Rates and Total Derivative with a real application problem. This video is made for http://B.Tech 1st Year, BSc Maths, MSc, JEE Advanced, CUET PG, and GATE Engineering Mathematics students who need step by step solutions for rate of change problems.
00:00 If x increases at the rate of 2cm/sec at the instant when x=3 cm and y=1cm at what rate must y be changing in order that the function 2xy-3x²y shall be neither increasing nor decreasing?
00:30 Total Derivative
05:00 Answer

If x is increasing at 2 cm per sec and at the instant x equals 3 cm and y equals 1 cm, we need to find at what rate y must be changing so that the function 2xy minus 3x²y is neither increasing nor decreasing. This means the total derivative of the function with respect to time must be zero.

00:00 If x increases at the rate of 2cm/sec at the instant when x=3 cm and y=1cm at what rate must y be changing in order that the function 2xy-3x²y shall be neither increasing nor decreasing?
We are given dx/dt = 2 cm per sec. At the given instant x = 3 cm, y = 1 cm.
Let F = 2xy minus 3x²y. For F to be neither increasing nor decreasing, dF/dt must be 0. We use total derivative to relate dx/dt and dy/dt.

00:30 Total Derivative
Formula: dF/dt = ∂F/∂x times dx/dt + ∂F/∂y times dy/dt
First find partial derivatives:
∂F/∂x = 2y minus 6xy
∂F/∂y = 2x minus 3x²

Now substitute x = 3, y = 1:
∂F/∂x = 2(1) minus 6(3)(1) = 2 minus 18 = minus 16
∂F/∂y = 2(3) minus 3(3)² = 6 minus 27 = minus 21

Given dx/dt = 2 cm per sec. Let dy/dt = k, which we need to find.

Set dF/dt = 0:
0 = minus 16 times 2 + minus 21 times k
0 = minus 32 minus 21k
21k = minus 32
k = minus 32 over 21 cm per sec

05:00 Answer
dy/dt = minus 32 over 21 cm per sec approximately minus 1.524 cm per sec
The negative sign means y must be decreasing at this rate so that the function remains constant

22/07/2026

426 | Total Derivative Solved | If u=x²+y², x=at², y=2at Find du/dt | Engineering Maths
Partial DIFFERENTIATION playlist - https://youtube.com/playlist?list=PLT2xRiy9TL2gDydKo8lzx-dQLLm3IFG8A&si=JBJbDytucgcuOf_h

Notes- https://youtube.com/shorts/09NmFTgSHh8?si=x50inz6L5miN4VAP
Learn Total Differentiation and Chain Rule with a complete solved example. This video is made for http://B.Tech 1st Year, BSc Maths, MSc, JEE Advanced, CUET PG, and GATE Engineering Mathematics students who need step by step solutions for differentiation of composite functions.
00:00 If u=x²+y², x=at², y=2at, find du/dt
00:30 Total Differentiation by Chain Rule
02:00 By Total Differentiation Formula
04:00 du/dt

If you are asked to find du/dt when u = x² plus y², x = at², y = 2at, this lecture explains two methods. First by direct substitution, second by total differentiation formula using chain rule. Both methods give the same answer and are important for university and competitive exams.

00:00 If u=x²+y², x=at², y=2at, find du/dt
We have u as a function of x and y, and both x and y are functions of t. So u is ultimately a function of t only. We will find du/dt using chain rule and also by substituting x and y first and then differentiating.

00:30 Total Differentiation by Chain Rule
Formula: du/dt = ∂u/∂x times dx/dt + ∂u/∂y times dy/dt
First find partial derivatives:
∂u/∂x = 2x
∂u/∂y = 2y

Now find derivatives of x and y with respect to t:
x = at² so dx/dt = 2at
y = 2at so dy/dt = 2a

02:00 By Total Differentiation Formula
Substitute in the formula:
du/dt = 2x times 2at + 2y times 2a
= 4atx + 4ay

Now put x = at² and y = 2at:
du/dt = 4at times at² + 4a times 2at
= 4a²t³ + 8a²t
= 4a²t(t² + 2)

04:00 du/dt by Direct Substitution Method
Method 2: Put x and y in u first.
u = x² + y² = (at²)² + (2at)² = a²t⁴ + 4a²t²
Now differentiate w.r.t t:
du/dt = 4a²t³ + 8a²t = 4a²t(t² + 2)
Both methods give the same result.

What you will gain from this video
1. Complete solution of total derivative using chain rule
2. How to handle parametric functions x=at² and y=2at
3. Comparison of substitution m

21/07/2026

425 | Total Differentiation Solved | If u=sin(x²+y²), a²x²+b²y²=c² Find du/dx | Engineering Maths

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