Derivatives
A derivative is the rate of change of a quantity y with respect to another quantity x. A derivative is also termed the differential coefficient of y with respect to x. Differentiation is the process of finding the derivative of a function.
What Are Derivatives?
If f(x) is a function differentiable in an interval [a,b], at every point of the interval the derivative of the function exists finitely and it is unique. There exists a new function g: [a,b] → R, such that, x ∈ [a,b], g(x) = f'(x). This f'(x) is the first derivative of the function f(x), that is denoted by \(\dfrac{df(x)}{dx}\) or Df(x) or f'(x).
Consider a curve of function f(x). Consider two points (x, f(x)) and ((x+ h), f(x+h)). Then the slope of the secant line through these points is given by \( \dfrac{f(x+h) f(x)}{x+h x}= \dfrac{f(x+h) f(x)}{h}\). To get the actual slope of the point, we use the limiting process of finding the derivative or slope of the second point by a variable 'h'. We try to calculate the ratio when h approaches 0. As h approaches 0, the second point approaches the original point and the secant line becomes the tangent line.
Interpretation of Derivatives
The derivative is a function that is geometrically defined as the slope of the line tangent to the curve at any point. If f is differentiable and continuous at [a,b], then f'(x) = \(\lim _{h \rightarrow 0} \dfrac{f(x+h) f(x)}{h}\). This change in h is infinitely very small. We denote it by Δx. Then change in the original function f(x) is also small, denoted by Δy. The derivative so obtained by applying the limits is also defined as the instantaneous rate of change of a function with respect to a variable.
f'(x) = \(\dfrac{Δy}{Δx} =\lim _{\delta x \rightarrow 0} \dfrac{f(x+\delta x) f(x)}{\delta x}\).
If the derivative > 0, then the curve is increasing, and if the derivative < 0, then the curve is decreasing. The derivative at any stationary point = 0 which implies that the function is neither increasing nor decreasing.
How To Find The Derivatives?
Derivatives are obtained by applying the limits as per the first principle of differentiation that we obtained as the definition of a derivative. Let f(x) = 4x^{2 }+ 3
\( f'(x) =\lim _{\delta x \rightarrow 0} \dfrac{f(x+\delta x) f(x)}{\delta x}\\ = \lim _{\delta x \rightarrow 0} \dfrac{(4(x+\delta x)^2 + 3)(4x^2 + 3)}{\delta x}\\=\lim _{\delta x \rightarrow 0} \dfrac{(4x^2 + 4\delta x^2 + 8x \delta x + 3)(4x^2 + 3)}{\delta x}\\=\lim _{\delta x \rightarrow 0} \dfrac{(4x^2 + 4\delta x^2+ 8x \delta x + 34x^2  3)}{\delta x}\\=\lim _{\delta x \rightarrow 0} \dfrac{4 \delta x^2 + 8x \delta x}{\delta x}\\= \lim _{\delta x \rightarrow 0} 4\delta x+8x\)
f'(x) = 8x
Therefore the first derivative of 4x^{2} + 3 = 8x.
Thus the first derivative of an algebraic function is derived using the definition of the derivative.
Derivatives of Elementary Functions
The three basic derivatives of the algebraic, logarithmic/ exponential and trigonometric functions are derived from the fundamental principle of differentiation and are used as standard derivative formulas. They are as follows:
 If y = ln\(_e\) x, then dy/dx = 1/x
 if y = log\(_a\) x, then dy/dx = 1/[(log a) x]
 If y = a^{ x }, dy/dx= a^{x }log a
Derivatives of Trigonometric Functions
 If y = sin x, y' = cos x
 If y = cos x, y' = sin x
 If y = tan x, y' = sec^{2} x
 If y = sec x, y' = sec x tan x
 If y = cosec x, y' = cosec x cot x
 If y = cot x, y' = cosec^{2} x
Derivatives of Inverse Trigonometric Functions
 If y = sin^{1} x, y' = \(\dfrac{1}{\sqrt{(1x^2)}}\)
 If y = cos^{1} x, y' = \(\dfrac{1}{\sqrt{(1x^2)}}\)
 If y = tan^{1} x, y' = \(\dfrac{1}{(1+x^2)}\)
 If y = cot^{1} x, y' =\(\dfrac{1}{(1+x^2)}\)
 If y = sec^{1} x, y' = \(\dfrac{1}{x \sqrt{(x^2 1)}}\)
 If y = cosec^{1} x, y' = \(\dfrac{1}{x \sqrt{(x^2 1)}}\)
Derivatives of Composite Functions
If f and g are the derivative functions in their domain, then f(g(x) is also differentiable. This is known as the chain rule of differentiation used for composite functions. (fog)'(x) = f'[(g(x)] g'(x). This also can be written as y = f(u) and t = g(x) , then dy/dx = dy/ dt . dt/dx
For example, consider y = tan^{2}x. This is a composite function. By the chain rule, y = u^{2 }, where u = tan x
du/dt = 2u
dy/du = d/dx .tan x = sec ^{2} x
dy/dx = dy/du . du/dx
= 2 u . sec ^{2} x
= 2 tan x. sec ^{2} x
Fundamental Rules of Derivatives
The following are the fundamental rules of derivatives. Let us discuss them in detail.
Power Rule: The power rule of derivatives states that if a function is an algebraic expression raised to any power, say n, then the derivative has a power 1 less than the original function. If y = x^{n }, where n > 0. Then dy/dx = n x ^{n1 }. Example: x^{5 }= 5x^{4}
Sum Rule: The sum rule of derivatives states that if a derivative is the sum/difference of two functions, then it is equal to the sum/difference of its derivatives. dy/dx [u(x) ± v(x)]= du/dx ± dv/dx.
Product Rule: The product rule of derivatives states that if a function is a product of two functions, then its derivative is the derivative of the second function multiplied by the first function added to the derivative of the first function multiplied by the second function. dy/dx [u(x) × v(x)] = u.dv/dx + v.du/dx. If y = x^{5 }e^{x }, we have y' = x^{5 }e^{x }+ e^{x }. 5x^{4 }= e^{x }(x^{5 }+ 5x^{4})
Quotient Rule: The quotient rule of derivatives states that if a function is of the form u(x)/v(x), then the derivative is the difference between the second function × derivative of the first function and the first function × derivative of the second function divided by the square of the second function. dy/dx [u(x) ÷ v(x)]= (v.du/dx u.dv/dx)/ v^{2 }
Constant multiple Rule: The constant multiple rule of derivatives states that if a function is the derivative multiplied by a constant then it is equal to the constant multiplied by its derivative d/dx [c(f(x)] = c. d/dx f(x), where c ≠ 0, and c is a constant. d/dx .5x^{2 }= 5. d/dx .x^{2 }= 5. 2x = 10 x.
Constant Rule: The constant rule of derivatives states that the derivative of any constant is 0. If y = k, where k is a constant, then dy/dx = 0. Suppose y = 4, y' = 0
Derivatives of Implicit Functions
In equations where y as a function of x cannot be explicitly defined by the variables x and y, we use implicit differentiation. If f(x,y) =0 , then differentiate with respect to x and group the terms containing dy/dx at one side and then solve for dy/dx.
For example, 2x + y = 12
d/dx(2x + y)= 0
2 + dy/dx = 0
dy/dx = 2
Parametric Derivatives
In a function, we may have the dependent variables x and y which are dependent on the third independent variable. If x = f(t) and y = g(t), then derivative is calculated as dy/dx = f'(x)/g'(x). Suppose, if x = 4 + t^{2 }and y = 4t^{2 }5t^{4 }, then let us find the parametric derivative.
dx/ dt = 2t and dy/dt = 8t 20t^{3}
dy/dx = (dy/dt)/(dx/dt)
dy/dx = (8t 20t^{3 })/2t
=2t(410t^{2 })/2t
dy/dx = 410t^{2}
Higherorder Derivatives
We can find the successive derivatives of a function and obtain the higherorder derivatives. If y is a function, then its first derivtive is dy/dx. The second derivative is \(\dfrac{d}{dx}.\dfrac{dy}{dx}\). The third derivative is \(\dfrac{d}{dx}.\dfrac{d^2y}{dx^2}\) and so on. Suppose y = 4x^{3 }, we get the successive derivatives as follows. y' = 12x^{2} , y'' = 24 x and y''' = 24, y''''= 0.
Partial Derivatives
If u=f(x,y) we can find the partial derivative of y keeping x as the constant or we can find the partial derivative of x by keeping y as the constant. Suppose f(x,y) = x^{3 }y^{2 }, the partial derivatives of the function are:
𝛿f/dx(x^{3 }y^{2}) = 3x^{2}y and
𝛿f/dy(x^{3 }y^{2}) = x^{3 }2y
Important Notes
 A derivative of a function is the rate of change of one quantity over the other.
 Derivative of any continuous function that is differentiable at [a,b] is derived using the first principle of differentiation using the limits.
 If f(x) is given,
f'(x) = \(\dfrac{Δy}{Δx} =\lim _{\delta x \rightarrow 0} \dfrac{f(x+\delta x) f(x)}{\delta x}\)
☛Also Check:
Examples of Derivatives

Example 1. Find the derivative of the curve y = [(x+3) (x+2)]/x^{2} at the point (3,0).
Solution:
y = (x^{2 }+ 5x +6)/ x^{2 }= 1 + 5/x + 6/ x^{2}
The derivative of the curve is dy/dx = d/dx. 1+ d/dx . 5/x + d/dx. 6/ x^{2 }
dy/dx = d/dx. 1 + 5 .d/dx. x ^{1 }+ 6. d/dx. x^{}^{2 }
= 0 + 5 (1. x ^{2 }) + 6 . (2. x^{}^{3 }) (Using the power rule of derivatives)
=  5/x^{2 }12/x^{3}
The derivative of the curve at x = 3 is dy/dx =  5/3^{2 }12/3^{3 }
=  5/9  12 / 27
= 27/27 = 1.
Answer: The derivative of the curve y = [(x+3) (x+2)]/x^{2} at the point (3,0) is 1.

Example 2. Suppose the position of an object after t hours is given by g(x) = x/(x+1), then determine if the object is moving to the right or left at t= 10 hours.
Solution:
The derivative of the given function is g'(x) = d/dt . [x/(x+1)]
[(x+1). d/dx.x x .d/dx. (x+1)] / (x+1)^{2 }[Applying the quotient rule of derivatives]
[(x+1) . 1  x. d/dx. (x+1)] / (x+1) ^{2}
[(x+1) . 1  x. d/dx(x). d/dx(1)] / (x+1) ^{2 }[Applying the sum rule of derivatives]
[(x+1) x(1+0)]/ (x+1) ^{2}
Thus g'(x) = 1/(x+1)^{2 }
To determine if the object is moving to the right or left, we need to find the derivative at t = 10.
g'(10) = 1/ 11^{2 }= 1/121
Since the derivative is positive, the velocity function is increasing.Answer: The velocity function is increasing and the object is moving to the right.

Example 3. What is the equation of the tangent line to f(x) = x^{2 }+ 5x at x = 4?
Solution:
Given f(x) = x^{2 }+ 5x
f(4) = 4^{2 }+ 5(4)= 16 + 20 = 36
The derivative of f(x) is the slope of the tangent line.
f'(x) = d/dx(x^{2 }+ 5x ) = 2x + 5
f'(4) = 2(4) +5 = 13
m =13
Thus the equation of the tangent at (4, 36) is given by the slope formula: y y\(_1\) = m (x  x\(_1\)
y 36 = 13( x  4)
y  36 = 13x  52
y = 13x 52 + 36
y = 13x 16
Answer: The equation of the tangent line to f(x) = x^{2} + 5x at x = 4 is y = 13x 16
FAQs on Derivatives
What Are Derivatives?
A derivative is the instantaneous rate of change of a quantity y with respect to another quantity x. Differentiation is the process of finding the derivative of a function. A derivative is also defined as the slope of a tangent of the curve at a point.
Define Derivative.
A derivative of f(x) at x = a is given by f(x) = \(\lim_{x \overrightarrow a} \dfrac{f(x)f(a)}{xa}\). It is the slope of the tangent to the function f(x). It is also called as the differential quotient.
How Do You Find Derivatives?
The derivatives of functions are found using the definition of derivative from the first fundamental principle of differentiation. If f(x) is a given function, its derivative is obtained using f'(x) = \(\dfrac{Δy}{Δx} =\lim _{\delta x \rightarrow 0} \dfrac{f(x+\delta x) f(x)}{\delta x}\). A lot of other substitution techniques, and rules are used to find the derivatives.
What Are The Three Basic Derivatives?
The three basic derivatives are those of the algebraic functions, trigonometric functions, and exponential functions. For example, if y = 3x, y' = 3, if y = cos x, y' = sin x and if y = e^{x }, y' = e^{x}
What is Derivative Formula?
The basic formulas that are derived from the fundamental principle of differentiation are the derivative formulas. We use them as standard formulas to find the derivatives of algebraic, trigonometric, and exponential functions.
What Are the Application of Derivatives in Real Life?
The rate of change of a function with respect to another quantity is the derivative. To check if a function is increasing or decreasing, to find the equation of the tangent/normal, to find the maximum and minimum values from a graph, to find the displacementmotion problems, to find velocity given displacement, to find the acceleration given displacement and so on.
What is a Derivative Example?
Speed is the instant rate of change of the distance taken by an object at a particular time. The first derivative of the displacement of an object is its velocity. The second derivative of displacement is the object's acceleration. The third derivative of the displacement is the object's jerk and so on.
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