In mathematics, differential calculus is used to determine the slope of the tangent line. Differential calculus is usually used to find the derivative of the given function according to the corresponding variables of the function.
The function can be single variable f(x), double variable f(x, y), or three variable f(x, y, z). The reverse process of this type of calculus is said to be the integral calculus. This post will discuss the definition, types, and rules of differential calculus, along with many examples.
What is differential calculus?
Differential calculus is a general branch of mathematics concerned with studying the instantaneous rate of change of single, double, and multivariable functions concerning independent variables.
The function of differential calculus could be linear, constant, polynomial, exponential, logarithmic, trigonometric, quadric, etc. The independent variables can be x, y, z, u, v, w, t, etc. and the notation of derivatives according to the variables are d/dx, d/dy, d/dz, d/du, d/dv, d/dw, etc.
Types of differential calculus
Differential calculus is of various types. Let’s discuss them briefly.
1. Explicit differentiation
The explicit derivative is a main and commonly used type of differential calculus to differentiate the one-variable function. For example, the derivative of 3t concerning t must be 3. It is denoted by d/dx, d/dy, d/dz, d/du, d/dv, d/dw, etc.
2. Implicit differentiation
Implicit differentiation is another commonly used type of differential calculus to solve the differential of implicit functions or equations. In this type of differential calculus, the function must have two variables, and the dependent variable must not be considered constant concerning the independent variable.
For example, if the term y is a dependent variable and x is an independent variable, then the differential of y concerning t must be y’ or dy/dt.
3. Partial derivative
This differential calculus solves the multivariable functions according to the respective variables. It differentiates the functions partially, i.e., solves the given multivariable functions concerning all the variables one by one.
It is denoted by, ∂/∂t [f (u, v, t)], ∂/∂u [f (u, v, t)], ∂/∂v [f (u, v, t)],
4. Directional derivative
The directional derivative is also a kind of differential calculus used to find the direction of the function by using the multivalued functions and the vector. It takes the partial derivatives known as gradient and normalizes the vector.
It is denoted by u * ∂f (t, w)/∂t.
Rules of differential calculus
Here are some rules of differential calculus.
1. Sum rule
d/dt [f(t) + h(t)] = d/dt [f(t)] + d/dt [h(t)]
2. Constant rule
d/dt [U] = 0, where U is any constant
3. Difference rule
d/dt [f(t) – h(t)] = d/dt [f(t)] – d/dt [h(t)]
4. Power rule
d/dt [fn(t)] = n * fn-1(t) * d/dt [f(t)]
5. Product rule
d/dt [f(t) x h(t)] = h(t) d/dt [f(t)] + f(t) d/dt [h(t)]
6. Quotient rule
d/dt [f(t) / h(t)] = 1/ (h(t))2 [h(t) d/dt [f(t)] – f(t) d/dt [h(t)]
7. Constant function rule
d/dt [U * f(t)] = U * d/dt [f(t)]
Examples of differential calculus
Here are some examples of differential calculus
Example-1
Find the differential of 10t5 + 4t3 – 8t2 + cos(t) + 4t + 6 with respect to “t”.
Solution
Step-1:Write the given differential function.
10t5 + 4t3 – 8t2 + cos(t) + 4t + 6
Step-2:Now write the function according to the notation of the single variable function.
d/dt [10t5 + 4t3 – 8t2 + cos(t) + 4t + 6]
Step-3:Now apply the single variable differential notation separately to each function with the help of the sum and difference rule of differential calculus.
d/dt [10t5 + 4t3 – 8t2 + cos(t) + 4t + 6] = d/dt [10t5] + d/dt [4t3] – d/dt [8t2] + d/dt [cos(t)] + d/dt [4t] + d/dt [6]
Step-4:Now use the constant function rule of differential calculus.
d/dt [10t5 + 4t3 – 8t2 + cos(t) + 4t + 6] = 10 d/dt [t5] + 4 d/dt [t3] – 8d/dt [t2] + d/dt [cos(t)] + 4 d/dt [t] + d/dt [6]
Step-5:Now find the differential of the above expression with the help of power, trigonometry, and the constant rule of differential calculus.
d/dt [10t5 + 4t3 – 8t2 + cos(t) + 4t + 6] = 10 d/dt [t5] + 4 d/dt [t3] – 8d/dt [t2] + d/dt [cos(t)] + 4 d/dt [t] + d/dt [6]
= 10 [5t5-1] + 4 [3t3-1] – 8 [2t2-1] + [-sin(t)] + 4 [t1-1] + [0]
= 10 [5t4] + 4 [3t2] – 8 [2t1] + [-sin(t)] + 4 [t0] + [0]
= 10 [5t4] + 4 [3t2] – 8 [2t] + [-sin(t)] + 4 [1] + [0]
= 10 [5t4] + 4 [3t2] – 8 [2t] – [sin(t)] + 4 [1] + [0]
= 50t4 + 12t2 – 16t – sin(t) + 4
To avoid such a large number of steps to get the result, you can use a differentiate calculator to get the result with steps in a fraction of seconds. Follow the below steps to calculate derivatives with steps using a calculator.
Step I: First, write the differential function into the required input box.
Step II: Choose the independent variable of the function.
Step III: Type the order of derivative.
Step IV: Press the calculate button.
Step V: The result with steps of the function will show below the calculate button.
Example-2:
Find the differential of,f(t, y) = 4ty2 + 12y3 – 11y + 3t + 4, g(t, y) = (2t4 * 3t2) + 7t – 4y2, with respect to “t”.
Solution
Step-1:Write the given differential function f(u, y) = g(u, y)
4ty2 + 12y3 – 11y + 3t + 4 = (2t4 * 3t2) + 7t – 4y2
Step-2:Now write the function according to the notation of the single variable function on both sides of the equation.
d/dt [4ty2 + 12y3 – 11y + 3t + 4] = d/dt [(2t4 * 3t2) + 7t – 4y2]
Step-3:Now apply the single variable differential notation separately to each function with the help of the sum, product, and difference rule of differential calculus.
d/dt [4ty2] + d/dt [12y3] – d/dt [11y] + d/dt [3t] + d/dt [4] = d/dt [(2t4 * 3t2)] + d/dt [7t] – d/dt [4y2]
y2 d/dt [4t] + 4t d/dt [4y2] + d/dt [12y3] – d/dt [11y] + d/dt [3t] + d/dt [4] = 3t2 d/dt [2t4] + 2t4 d/dt [3t2] + d/dt [7t] – d/dt [4y2]
Step-4:Now use the constant function rule of differential calculus.
4y2 d/dt [t] + 16t d/dt [y2] + 12 d/dt [y3] – 11 d/dt [y] + 3 d/dt [t] + d/dt [4] = 6t2 d/dt [t4] + 6t4 d/dt [t2] + 7 d/dt [t] – 4 d/dt [y2]
Step-5:Now find the differential of the above expression with the help of power and the constant rule of differential calculus.
4y2 [t1-1] + 16t [2y2-1 dy/dt] + 12 [3y3-1 dy/dt] – 11 d/dt [y] + 3 [t1-1] + [0] = 6t2 [4t4-1] + 6t4 [2t2-1] + 7 [t1-1] – 4 [2y2-1 dy/dt]
4y2 [t0] + 16t [2y1 dy/dt] + 12 [3y2 dy/dt] – 11 d/dt [y] + 3 [t0] + [0] = 6t2 [4t3] + 6t4 [2t1] + 7 [t0] – 4 [2y1 dy/dt]
4y2 [1] + 16t [2ydy/dt] + 12 [3y2 dy/dt] – 11 d/dt [y] + 3 [1] + [0] = 6t2 [4t3] + 6t4 [2t] + 7 [1] – 4 [2ydy/dt]
4y2 [1] + 16t [2ydy/dt] + 12 [3y2 dy/dt] – 11 d/dt [y] + 3 [1] + [0] = 6t2 [4t3] + 6t4 [2t] + 7 [1] – 4 [2ydy/dt]
4y2 + 16t [2ydy/dt] + 12 [3y2 dy/dt] – 11 d/dt [y] + 3 = 6t2 [4t3] + 6t4 [2t] + 7 – 4 [2ydy/dt]
4y2 + 16t [2ydy/dt] + 12 [3y2 dy/dt] – 11 d/dt [y] + 3 = 6t2 [4t3] + 6t4 [2t] + 7 – 4 [2ydy/dt]
4y2 + 32tydy/dt + 36y2 dy/dt – 11 dy/dt + 3 = 24t5 + 12t5 + 7 – 8ydy/dt
Step-6:Now separate the dy/dt terms.
32tydy/dt + 36y2 dy/dt – 11 dy/dt + 8ydy/dt = 24t5 + 12t5 + 7 – 4y2 – 3
[32ty+ 36y2 – 11 + 8y]dy/dt = 24t5 + 12t5 + 7 – 4y2 – 3
dy/dt = 24t5 + 12t5 + 7 – 4y2 – 3] / [32ty+ 36y2 – 11 + 8y]
Summary:
This post has discussed all the basics of differential calculus and many solved examples. After reading the above post, you can easily solve differential calculus problems.
