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)] = lim_{h}_{→0} 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) |
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’ [e^{t}]
= e^{t} |
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) = 2t^{4} + 4t^{3} – 2t
/ 5t^{2} + 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) = 2t^{4} + 4t^{3} – 2t
/ 5t^{2} + 20t
Corresponding variable = t
d/dt f(t) = d/dt [2t^{4} + 4t^{3}
– 2t / 5t^{2} + 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 [2t^{4} + 4t^{3} – 2t
/ 5t^{2} + 20t] = d/dt [2t^{4}] + d/dt [4t^{3}] – d/dt
[2t / 5t^{2}] + d/dt [20t]
Step 3: Now apply the quotient law of differential calculus.
d/dt [2t^{4} + 4t^{3} – 2t
/ 5t^{2} + 20t] = d/dt [2t^{4}] + d/dt [4t^{3}] –
[1/(2t)^{2} [2t d/dt [5t^{2}] – 5t^{2} d/dt [2t]] +
d/dt [20t]
d/dt [2t^{4} + 4t^{3} – 2t
/ 5t^{2} + 20t] = d/dt [2t^{4}] + d/dt [4t^{3}] –
1/(2t)^{2} [2t d/dt [5t^{2}] + 1/(2t)^{2} [5t^{2}
d/dt [2t]] + d/dt [20t]
Step 3: Now take the constant
coefficients outside the differential notation.
d/dt [2t^{4} + 4t^{3} – 2t
/ 5t^{2} + 20t] = 2d/dt [t^{4}] + 4d/dt [t^{3}] –
1/(2t)^{2} [2t d/dt [5t^{2}] + 1/(2t)^{2} [5t^{2}
d/dt [2t]] + 20d/dt [t]
Step 4: Now differentiate the
above expression with the help of the power rule.
= 2[4t^{4-1}] + 4 [3t^{3-1}]
– 1/(2t)^{2} [2t [10t^{2-1}] + 1/(2t)^{2} [5t^{2}
[2t^{1-1}]] + 20 [t^{1-1}]
= 2[4t^{3}] + 4 [3t^{2}] –
1/(2t)^{2} [2t [10t^{1}] + 1/(2t)^{2} [5t^{2}
[2t^{0}]] + 20 [t^{0}]
= 2[4t^{3}] + 4 [3t^{2}] –
1/(2t)^{2} [2t [10t] + 1/(2t)^{2} [5t^{2} [2(1)]] + 20
[1]
= 2[4t^{3}] + 4 [3t^{2}] –
1/(2t)^{2} [2t [10t] + 1/(2t)^{2} [5t^{2} [2]] + 20
= 8t^{3} + 12t^{2} – 1/(2t)^{2}
[20t^{2}] + 1/(2t)^{2} [10t^{2}] + 20
= 8t^{3} + 12t^{2} – 20t^{2}/2t^{2}
+ 10t^{2}/2t^{2} + 20
= 8t^{3} + 12t^{2} – 10 + 5
+ 20
= 8t^{3} + 12t^{2} – 5 + 20
= 8t^{3} + 12t^{2} + 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) = 2r^{2} – 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)] = lim_{h}_{→0} 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) = 2r^{2} – 6r + 14
f(r) = 2(r + h)^{2} – 6(r + h) + 14
f(r) = 2(r^{2} + 2hr + h^{2})
– 6(r + h) + 14
f(r) = 2r^{2} + 4hr + 2h^{2}
– 6r – 6h + 14
Step 3: Now substitute the values in the general expression of the first
principle method
d/dr [2r^{2} – 6r + 14] = lim_{h}_{→0} [2r^{2} + 4hr + 2h^{2}
– 6r – 6h + 14] – [2r^{2} – 6r + 14] / h
d/dr [2r^{2} – 6r + 14] = lim_{h}_{→0} [2r^{2} + 4hr + 2h^{2}
– 6r – 6h + 14 – 2r^{2} + 6r – 14] / h
d/dr [2r^{2} – 6r + 14] = lim_{h}_{→0} [4hr + 2h^{2} – 6h] / h
d/dr [2r^{2} – 6r + 14] = lim_{h}_{→0} [4r + 2h – 6]
d/dr [2r^{2} – 6r + 14] = lim_{h}_{→0} [4r] + lim_{h}_{→0} [2h] – lim_{h}_{→0} [6]
d/dr [2r^{2} – 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.