Just as algebra introduces students to new ways of thinking about arithmetic problems (by way of variables, equations, functions and graphs), calculus may be seen as introducing new ways of thinking about algebra problems.
A branch of mathematics, algebra calculus mainly deals with derivatives, integrals, limits, infinite series, differential equations, and functions, is called calculus. It is used for computing rate of change of velocity and rate of change of acceleration, the slope of a curve, and optimization
Types of Calculus and Problems in Algebra Problems:
Calculus can be classified into two major types as follows
Differential calculus
Integral calculus
Ex 1: Compute the values of Δy and dy if y = f(x) = x^3 + x^2 − 2x + 1 where x changes (i) from 2 to 2.05 and (ii) from 2 to 2.01
Sol : (i) We have f(2) = 2^3 + 2^2 − 2(2) + 1 = 9
f(2.05) = (2.05)^3 + (2.05)^2 − 2(2.05) + 1 = 9.717625.
And Δy = f(2.05) − f(2) = 0.717625.
In general dy = f ′(x) dx = (3x^2 + 2x − 2)dx
When x = 2, dx = Δx = 0.05 and dy = [(3(2)^2+2(2)−2] 0.05 = 0.7
(ii) f(2.01) = (2.01)^3 − (2.01)^2 − 2(2.01) + 1 = 9.140701
Δy = f(2.01) − f(2) = 0.140701
When dx = Δx = 0.01, dy = [3(2)^2 + 2(2) − 2]0.01 = 0.14
Remark : The approximation Δy ≈ dy becomes better as Δx becomes smaller . Also dy was easier than to compute Δy. For more complicatedfunctions it may be impossible to compute Δy exactly. In such cases theapproximation by calculus is especially useful.
Example 2 Algebra Calculus Problems:
Ex 2 : Use differentials to find an approximate value for
`root(3)65` .
Let y = f(x) =`3sqrtx`=x.^1/3 Then dy =1/3x ^-2/3 dx
Since f(64) = 4. We take x = 64 and dx = Δx = 1
This gives dy =1/3 (64) ^−2/3 (1)
=1/3(16) = 1/48
365 = f(64 + 1) ≈ f(64) + dy = 4 +1/48
≈ 4.021
Practice Problems
Pro 1: If f(x) = x^4, then find f'(x)
Sol : f(x) = x^4
f'(x) = 4x^3
Pro 2: If f(x) = 5x3^ + 2x^2, then find f'(x).
Sol : f(x) = 5x^3 + 2x^2
f'(x) = 15x^2 + 4x
Pro 3: x7 dx = (x^7+1) / (7 + 1) + c
Sol : = x^8/8 + c
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