How to evaluate the differential of the functions? Explained with solutions

 Mathematics is a vast subject that is used almost in every main field of education and daily life calculations. There are further subtypes of mathematics such as calculus, geometry, algebra, trigonometry, matrices, sets, etc.

All these branches of mathematics are widely used to calculate various kinds of problems. There are two main sub-branches of calculus first is integral and the other is differential. In this post, we are going to explain one of two branches of calculus which is the differentiation of the function. we will also learn how to differentiate the functions. 

What is differential calculus?

In calculus, the instantaneous rate of change of the function according to the corresponding independent variable is known as the differential. Differential calculus is frequently used to calculate the slope of the tangent line. It is denoted by f’(t) or d/dt.

The process of finding the differential of the function with respect to the independent variable is known as differentiation. The function can be differentiated easily either by using the laws of differentiation or by using the first principle method.

The limit calculus is widely used to define and calculate the differential of the function with the help of the first principle method. The differential can be taken of the algebraic, trigonometric, logarithmic, and any other form of functions but the differential of the constant function always give zero output.

Formula of first principle method

Here is the general expression of differentiating the function with the help of limit calculus.

d/dt [f(t)] = limh0 f(t + h) – f(t) / h

the above formula will help you to calculate the differential of the function easily.

Types of differentiation

There are further many types of differentiation. Here are the well-known types of differentiation.

1.     Explicit differentiation

2.     Implicit differentiation

3.     Partial differentiation

The above sub-types of differentiation are frequently used to differentiate the single variable function, double variable function, and multivariable function. The term explicit differentiation is used to differentiate the single variable function.

The double variable function can be evaluated with the help of implicit differentiation. While partial differentiation is used to evaluate the multi-variable function. The differential of the function can be evaluated in various ways.

Laws of differential calculus

The laws of differential calculus play a vital role in differentiating the functions. Here are some well-known laws of differential calculus.

Laws name

Laws

Trigonometry Law

f’ [sin(t)] = cos(t)
f’ [cos(t)] = -sin(t)
f’ [tan(t)] = sec2(t)

Power Law

f’ [f(t)]n = n [f(t)]n-1 * [f’(t)]

Quotient Law

f’ [f(t) / g(t)] = 1/[g(t)]2 [g(t) * [f’(t)] - f(t) * [g’(t)]]

Product Law

f’ [f(t) * g(t)] = g(t) * [f’(t)] + f(t) * [g’(t)]

Constant Law

f’ [L] = 0, where L is any constant

Difference Law

f’ [f(t) - g(t)] = [f’(t)] - [g’(t)]

Sum Law

d/dt [f(t) + g(t)] = [f’(t)] + [g’(t)]

Exponential Law

f’ [et] = et

Constant function Law

f’ [L * f(t)] = L [f’(t)]

 

How to evaluate the differential of the function?

the first principle method and the laws of differential calculus are very helpful to evaluate the differential of the function. Let us take a few examples solved by the first principle and laws of differentiation.

Example 1: By using the laws

Evaluate the differential of the given function with respect to “t”.

f(t) = 2t4 + 4t3 – 2t / 5t2 + 20t

Solution

Step 1: First of all, take the given function (f(t)) and corresponding variable after that apply the differential notation to it.

f(t) = 2t4 + 4t3 – 2t / 5t2 + 20t

Corresponding variable = t

d/dt f(t) = d/dt [2t4 + 4t3 – 2t / 5t2 + 20t]

Step 2: Now use the sum and difference law of differentiation to apply the differential notation to each term of the function separately.

d/dt [2t4 + 4t3 – 2t / 5t2 + 20t] = d/dt [2t4] + d/dt [4t3] – d/dt [2t / 5t2] + d/dt [20t]

Step 3: Now apply the quotient law of differential calculus.

d/dt [2t4 + 4t3 – 2t / 5t2 + 20t] = d/dt [2t4] + d/dt [4t3] – [1/(2t)2 [2t d/dt [5t2] – 5t2 d/dt [2t]] + d/dt [20t]

d/dt [2t4 + 4t3 – 2t / 5t2 + 20t] = d/dt [2t4] + d/dt [4t3] – 1/(2t)2 [2t d/dt [5t2] + 1/(2t)2 [5t2 d/dt [2t]] + d/dt [20t]

Step 3: Now take the constant coefficients outside the differential notation.

d/dt [2t4 + 4t3 – 2t / 5t2 + 20t] = 2d/dt [t4] + 4d/dt [t3] – 1/(2t)2 [2t d/dt [5t2] + 1/(2t)2 [5t2 d/dt [2t]] + 20d/dt [t]

Step 4: Now differentiate the above expression with the help of the power rule.

= 2[4t4-1] + 4 [3t3-1] – 1/(2t)2 [2t [10t2-1] + 1/(2t)2 [5t2 [2t1-1]] + 20 [t1-1]

= 2[4t3] + 4 [3t2] – 1/(2t)2 [2t [10t1] + 1/(2t)2 [5t2 [2t0]] + 20 [t0]

= 2[4t3] + 4 [3t2] – 1/(2t)2 [2t [10t] + 1/(2t)2 [5t2 [2(1)]] + 20 [1]

= 2[4t3] + 4 [3t2] – 1/(2t)2 [2t [10t] + 1/(2t)2 [5t2 [2]] + 20

= 8t3 + 12t2 – 1/(2t)2 [20t2] + 1/(2t)2 [10t2] + 20

= 8t3 + 12t2 – 20t2/2t2 + 10t2/2t2 + 20

= 8t3 + 12t2 – 10 + 5 + 20

= 8t3 + 12t2 – 5 + 20

= 8t3 + 12t2 + 15

To ease up the larger calculations, a derivative calculator can be used. This calculator will give you the step-by-step solution to the given problem in a single click.




Follow the below steps to use this calculator.

1.     1.     Enter the function into the required input field.

2.     2. Select the corresponding variable.

3.    3.  Enter the number of derivatives you want to calculate such as 1 for the first derivative, 2 for the second derivative, etc.

4.    4.  Hit the calculate button. A solution with a step will show below the calculate button.

5.     5. Press the reset key to enter a new function. 

Example 2: By the first principle method

Evaluate the differential of the given function with respect to “r”.

f(r) = 2r2 – 6r + 14

Solution

Step 1: First of all, take the expression of the first principle method that is helpful in calculating the differential of the function.

d/dr [f(r)] = limh0 f(r + h) – f(r) / h

Step 2: Now take the f(w) and make f(r + h) with the help of the given function. 

f(r) = 2r2 – 6r + 14

f(r) = 2(r + h)2 – 6(r + h) + 14

f(r) = 2(r2 + 2hr + h2) – 6(r + h) + 14

f(r) = 2r2 + 4hr + 2h2 – 6r – 6h + 14

Step 3: Now substitute the values in the general expression of the first principle method

d/dr [2r2 – 6r + 14] = limh0 [2r2 + 4hr + 2h2 – 6r – 6h + 14] – [2r2 – 6r + 14] / h

d/dr [2r2 – 6r + 14] = limh0 [2r2 + 4hr + 2h2 – 6r – 6h + 14 – 2r2 + 6r – 14] / h

d/dr [2r2 – 6r + 14] = limh0 [4hr + 2h2 – 6h] / h

d/dr [2r2 – 6r + 14] = limh0 [4r + 2h – 6]

d/dr [2r2 – 6r + 14] = limh0 [4r] + limh0 [2h] – limh0 [6]

d/dr [2r2 – 6r + 14] = 4r + [2(0)] – 6 = 4r – 6

Final words

Now you can solve any problem of differential calculus by following the above post. As we have discussed all the basics of differential calculus along with its definition, types, rules, and solved examples. You can grab all the basics of this post just by learning it.

 

 

Previous Post Next Post