Calculus: Differentiation and Its Applications
What Is Differentiation?
Differentiation is a fundamental concept in calculus that deals with the rate of change of a function. In simple terms, it helps us find how a quantity is changing at any given point. The derivative of a function represents its slope or rate of change at a specific point.
The Derivative: Basic Definition
The derivative of a function f(x)f(x)f(x) is defined as the limit of the average rate of change of the function as the change in xxx approaches zero. It is written as:
f′(x)=limΔx→0f(x+Δx)−f(x)Δxf'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) – f(x)}{\Delta x}f′(x)=Δx→0limΔxf(x+Δx)−f(x)
The derivative f′(x)f'(x)f′(x) tells us how the value of f(x)f(x)f(x) changes as xxx changes.
Basic Differentiation Rules
Power Rule
The power rule is one of the most fundamental rules in differentiation. If f(x)=xnf(x) = x^nf(x)=xn, then:
f′(x)=n⋅xn−1f'(x) = n \cdot x^{n-1}f′(x)=n⋅xn−1
For example, if f(x)=x3f(x) = x^3f(x)=x3, then:
f′(x)=3x2f'(x) = 3x^2f′(x)=3×2
Constant Rule
If f(x)=cf(x) = cf(x)=c, where ccc is a constant, then:
f′(x)=0f'(x) = 0f′(x)=0
This is because the rate of change of a constant is always zero.
Sum and Difference Rule
If f(x)=g(x)+h(x)f(x) = g(x) + h(x)f(x)=g(x)+h(x), then:
f′(x)=g′(x)+h′(x)f'(x) = g'(x) + h'(x)f′(x)=g′(x)+h′(x)
Similarly, for subtraction:
f′(x)=g′(x)−h′(x)f'(x) = g'(x) – h'(x)f′(x)=g′(x)−h′(x)
Product Rule
For two functions u(x)u(x)u(x) and v(x)v(x)v(x), the product rule states:
f′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x)f′(x)=u′(x)v(x)+u(x)v′(x)
Quotient Rule
For the quotient of two functions u(x)u(x)u(x) and v(x)v(x)v(x), the quotient rule is:
f′(x)=v(x)u′(x)−u(x)v′(x)v(x)2f'(x) = \frac{v(x)u'(x) – u(x)v'(x)}{v(x)^2}f′(x)=v(x)2v(x)u′(x)−u(x)v′(x)
Applications of Differentiation
Finding Tangents to Curves
One of the most important applications of differentiation is finding the tangent to a curve at a specific point. The slope of the tangent line at any point on the curve is the value of the derivative at that point.
Example:
Find the equation of the tangent to the curve y=x2y = x^2y=x2 at the point x=3x = 3x=3.
- First, differentiate the function:
y′=2xy’ = 2xy′=2x
- Evaluate the derivative at x=3x = 3x=3:
y′=2(3)=6y’ = 2(3) = 6y′=2(3)=6
- The slope of the tangent is 666, and the point on the curve is (3,9)(3, 9)(3,9). Therefore, the equation of the tangent is:
y−9=6(x−3)y – 9 = 6(x – 3)y−9=6(x−3) y=6x−9y = 6x – 9y=6x−9
Velocity and Acceleration in Science
In Science, differentiation is used to calculate velocity and acceleration. If the position of an object is given by a function s(t)s(t)s(t), then the velocity v(t)v(t)v(t) is the first derivative of the position function with respect to time:
v(t)=dsdtv(t) = \frac{ds}{dt}v(t)=dtds
Similarly, acceleration a(t)a(t)a(t) is the derivative of the velocity:
a(t)=dvdta(t) = \frac{dv}{dt}a(t)=dtdv
Optimization Problems
Differentiation is widely used in optimization problems, where we need to find the maximum or minimum values of a function. To find the critical points (potential maxima or minima), we set the derivative equal to zero and solve for xxx.
Example Problem
Problem: Differentiate the following function and find the critical points:
f(x)=3×4−4×3+2×2−5f(x) = 3x^4 – 4x^3 + 2x^2 – 5f(x)=3×4−4×3+2×2−5
Solution:
To find f′(x)f'(x)f′(x), apply the power rule:
f′(x)=12×3−12×2+4xf'(x) = 12x^3 – 12x^2 + 4xf′(x)=12×3−12×2+4x
Now, set the derivative equal to zero to find the critical points:
12×3−12×2+4x=012x^3 – 12x^2 + 4x = 012×3−12×2+4x=0
Factor out 4x4x4x:
4x(3×2−3x+1)=04x(3x^2 – 3x + 1) = 04x(3×2−3x+1)=0
This gives us x=0x = 0x=0. To find the other critical points, solve 3×2−3x+1=03x^2 – 3x + 1 = 03×2−3x+1=0, which has no real solutions, so x=0x = 0x=0 is the only critical point.
Common Mistakes in Differentiation
- Forgetting to Use the Power Rule Correctly: Always subtract 1 from the exponent when using the power rule.
- Confusing the Sum and Product Rules: Be sure to apply the correct rule for sums and products of functions.
- Misinterpreting the Quotient Rule: The quotient rule involves both the numerator and denominator’s derivatives.
Practice Questions
- Differentiate f(x)=5×3+2×2−x+7f(x) = 5x^3 + 2x^2 – x + 7f(x)=5×3+2×2−x+7.
- Find the tangent equation to the curve y=x3−4×2+2xy = x^3 – 4x^2 + 2xy=x3−4×2+2x at x=2x = 2x=2.
- A particle moves along a straight line, and its displacement is given by s(t)=t2−4t+5s(t) = t^2 – 4t + 5s(t)=t2−4t+5. Find its velocity at t=3t = 3t=3.
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