What is differential calculus and how to find it?
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In mathematics, differential calculus is
one of the well-known branches of calculus that is widely used to find the
differential of the function. the differential of the function can be found
with respect to an independent variable.
This branch of calculus is also used to
find the slope of the tangent line. The differential function can be calculated
either by using rules of differentiation or the first principle method. In this
article, we will learn the basics of differential calculus along with examples.
Differential calculus
The differential
calculus is frequently used in mathematical analysis concerned with the
problem to determine the rate of change of function according to the
independent variable. Differential calculus is generally used to find the
differential of the functions and use them to evaluate the problems of
non-constant rate of change.
The exponential, constant, trigonometric,
polynomial, linear, logarithmic, or quadric functions are involved in
differential calculus.
Kinds of differential calculus
There are several kinds of derivatives in
calculus. Let us discuss some of them.
1.
Explicit differentiation
Explicit differentiation is the main kind
of differential calculus used to differentiate the single variable function
with respect to the independent variable. This kind of differential calculus is
denoted by f’(x) or d/dx. The independent variable could be x, y, z, u, v, w,
t, etc.
2.
Partial differentiation
Partial differentiation is that kind of differential
calculus that involves multivariable functions. It differentiates the function
with respect to any variable available in the function. the partial derivative
is denoted by ∂f(x, y, z)/∂x, ∂f(x, y, z)/∂y, and ∂f(x, y, z)/∂z.
3.
Implicit differentiation
Implicit differentiation is that type of differential
calculus that deals with the equation and implicit functions. The implicit
function is written as f(x, y) = g(x, y). The differentiation notation must be
applied on both sides of the equation.
This kind of differentiation determines the derivative of the dependent variable with respect to the independent variable without considering it as a constant. It is denoted by sdy/dx or y’(x).
Rules of
differential calculus
Here are some commonly used rules of differential
calculus.
1.
Sum rule: d/dw [g(w) + h(w)] = d/dw g(w)
+ d/dw h(w)
2.
Difference rule: d/dw [g(w) - h(w)] =
d/dw g(w) - d/dw h(w)
3.
Constant rule: d/dw [c] = 0, where c is
any constant
4.
Constant function rule: d/dw [c * f(w)]
= cd/dw [f(w)], where c is any constant
5.
Power rule: d/dw [f(w)] n =
n[f(w)] n-1
6.
Product rule: d/dw [g(w) * h(w)] = h(w)
* [d/dw g(w)] + g(w) * [d/dw h(w)]
7.
Quotient rule: d/dw [g(w) / h(w)] =
1/(h(w))2 [h(w) * [d/dw g(w)] - g(w) * [d/dw h(w)]]
8.
Chain rule: dy/dw= [dy/du * du/dw]
How to solve differential calculus problems?
Let us take a few examples to learn how to
solve the problems of differential calculus.
Example
1
Find the differential of 4w3 + 6w2 + cos(w) – 3w + 9 with respect to w.
Solution
Step 1:
First of all, apply the notation of differentiation to
the given function.
d/dw [4w3 + 6w2 +
cos(w) – 3w + 9]
Step 2:
Now write the differential notation to each function
separately by using the sum and difference rules of differential calculus.
d/dw [4w3 + 6w2 +
cos(w) – 3w + 9] = d/dw [4w3] + d/dw [6w2] + d/dw [cos(w)]
– d/dw [3w] + d/dw [9]
Step 3:
Use the constant function rule of differential calculus
and write the constant coefficients outside the differential notation.
d/dw [4w3 + 6w2 +
cos(w) – 3w + 9] = 4 d/dw [w3] + 6 d/dw [w2] + d/dw
[cos(w)] – 3 d/dw [w] + d/dw [9]
Step 4:
Differentiate the above expression with respect to “w”
with the help of power, trigonometric, and constant rules of
differentiation.
d/dw [4w3 + 6w2 +
cos(w) – 3w + 9] = 4 [3w3 – 1] + 6 [2w2 – 1] + [-sin(w) *
dw/dw] – 3 [w1 – 1] + [0]
= 4 [3w2] + 6 [2w1] +
[-sin(w) * dw/dw] – 3 [w0] + [0]
= 4 [3w2] + 6 [2w] + [-sin(w) * 1] – 3 [1] + [0]
= 12w2
+ 12w - sin(w) – 3
Use a differentiation
calculator to find the differential of single variable function to avoid
larger calculations.
Example 2: For implicit differentiation
Find the differential of 2wy – 3w2 + 2w – sin(y) = 2w2y + 4w – 24y + 2 w.r.t “w”.
Solution
Step 1: First of all, take
the given equation and write the differentiation notation on it.
2wy – 3w2 + 2w – sin(y) = 2w2y
+ 4w – 24y + 2
d/dw [2wy – 3w2 + 2w – sin(y)] =
d/dw [2w2y + 4w – 24y + 2]
Step 2:
Now write the differential notation to each function
separately by using the sum and difference rules of differential calculus.
d/dw [2wy] – d/dw [3w2] + d/dw
[2w] – d/dw [sin(y)] = d/dw [2w2y] + d/dw [4w] – d/dw [24y] + d/dw
[2]
Step 3:
Use the product rule of differential calculus.
2y d/dw [w] + 2w d/dw [y] – d/dw [3w2]
+ d/dw [2w] – d/dw [sin(y)] = 2y d/dw [w2] + 2w2 d/dw [y]
+ d/dw [4w] – d/dw [24y] + d/dw [2]
Step 4:
Differentiate the above expression with respect to “w”
with the help of power, trigonometric, and constant rules of
differentiation.
2y [w1 – 1] + 2w d/dw [y] – [3 *
2 w2 – 1] + [2 w1 – 1] – [cos(y) dy/dw] = 2y [2 w2 –
1] + 2w2 d/dw [y] + [4 w1 - 1] – d/dw [24y] + [0]
2y [w0] + 2w d/dw [y] – [3 * 2 w1]
+ [2 w0] – [cos(y) dy/dw] = 2y [2 w1] + 2w2
d/dw [y] + [4 w0] – d/dw [24y] + [0]
2y [1] + 2w d/dw [y] – [3 * 2 w] + [2 * 1]
– [cos(y) dy/dw] = 2y [2 w] + 2w2 d/dw [y] + [4 * 1] – d/dw [24y] +
[0]
2y + 2w d/dw [y] – [3 * 2 w] + [2] –
[cos(y) dy/dw] = 2y [2 w] + 2w2 d/dw [y] + [4] – d/dw [24y] + [0]
2y + 2w dy/dw – [6 w] + 2 – cos(y) dy/dw =
4wy + 2w2 dy/dw + 4 – 24 dy/dw
Step 5: Take the dy/dw terms
on the same side of the equation.
2w dy/dw – cos(y) dy/dw + 24 dy/dw - 2w2
dy/dw = 4wy + 6w - 2 + 4 – 2y
2w dy/dw – cos(y) dy/dw + 24 dy/dw - 2w2
dy/dw = 4wy + 6w + 2 – 2y
(2w – cos(y) + 24 - 2w2) dy/dw =
4wy + 6w + 2 – 2y
dy/dw = (4wy + 6w + 2 – 2y) / (2w – cos(y) + 24 - 2w2)
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