Introduction to derivative of quadratic functions:
In calculus the derivative is a measure of how a function changes as its input changes. The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. A quadratic equation is a polynomial equation of the second degree. The general form is ax2 + bx + c . The constants a, b, and c, are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term.
Source Wikipedia.
Derivative Formulas:
1. ` d / dx` (x n ) = n xn-1
2. ` d/dx` (ex) = ex
3. ` d/dx` `(uv)` = `u(dv)/(dx)` + `v(du)/(dx)`
4. .`d/dx` `(u/v)` = `(v(du)/(dx) - u(dv)/(dx))/ v^2`
5. `d/dx` (sin x) = cos x
6. `d/dx ` (cos x) = -sin x
7. `d/dx` (tan x) = sec2x
Derivatives of Quadratic Function Problems:
Derivatives of quadratic function problem 1:
Find the derivative of quadratic function (2x3 - 3x4 + x2) with respect to x.
Solution:
Given Quadratic function is (2x3 - 3x4 + x2)
` d/dx` (2x3 - 3x4 + x2) = `d/dx` (2x3) - ` d/dx` (3x4) + ` d/dx ` (x2) .
= `d/dx` (2x3) - `d/dx` (3x4) + `d/dx` (x2).
= (2 × 3) x(3-1) - (3 × 4) x(4-1) - 2x(2-1)
= 6 x2 - 12 x3 + 2 x
`d/dx` (2x3 - 3x4 + x2) = 6 x2 - 12 x3 + 2 x
Answer: The derivative of given quadratic function is `d/dx` (2x3 - 3x4 + x2) = 6x2 - 12 x3 + 2 x
Derivatives of quadratic function problem 2:
Find the derivative of quadratic function y = 3t2 + t3 - 30t. with respect to 't'.Is this topic free algebra help hard for you? Watch out for my coming posts.
Solution:
Given Quadratic function is y = 3t2 + t3 - 30t.
dy = (3 × 2)t(2 - 1) dt + (1 × 3)t(3 - 1) dt - 30t(1 - 1) dt.
= 6t dt + 3t2 dt - 30dt.
= (6t + 3t2 - 30) dt
`dy/(dt)` = (6t + 3t2 - 30)
Answer: The derivative of given quadratic function is `d/dt` (3 t2 + t3 - 30t) = 3t2 + 6t - 30.
Derivatives of quadratic function problem 3:
Determine the second derivative of given quadratic function f(x) = 15x2 + 10x + 5 with respect to x.
Solution:
Given quadratic function is, f(x) = 15x2 + 10x + 5
The First derivative of quadratic function is f'
f' =` (df )/ (dx)` = (15 × 2)x(2 - 1) + (10 × 1)x (1-1)
= 30x + 10
The Second derivative of quadratic function is, f''
` d^2/dx` f(x) = `d^2/dx` ( 15x2 + 10x + 5)
= `d/dx` (f')
= `d/dx` ( 30x + 10)
= ` d/dx` (30x) +` d/dx` (10)
= 30 `d/dx` (x) + `d/dx` (10)
= 30 + 0
= 30
Answer: The Second derivative of given quadratic function is 30
In calculus the derivative is a measure of how a function changes as its input changes. The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. A quadratic equation is a polynomial equation of the second degree. The general form is ax2 + bx + c . The constants a, b, and c, are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term.
Source Wikipedia.
Derivative Formulas:
1. ` d / dx` (x n ) = n xn-1
2. ` d/dx` (ex) = ex
3. ` d/dx` `(uv)` = `u(dv)/(dx)` + `v(du)/(dx)`
4. .`d/dx` `(u/v)` = `(v(du)/(dx) - u(dv)/(dx))/ v^2`
5. `d/dx` (sin x) = cos x
6. `d/dx ` (cos x) = -sin x
7. `d/dx` (tan x) = sec2x
Derivatives of Quadratic Function Problems:
Derivatives of quadratic function problem 1:
Find the derivative of quadratic function (2x3 - 3x4 + x2) with respect to x.
Solution:
Given Quadratic function is (2x3 - 3x4 + x2)
` d/dx` (2x3 - 3x4 + x2) = `d/dx` (2x3) - ` d/dx` (3x4) + ` d/dx ` (x2) .
= `d/dx` (2x3) - `d/dx` (3x4) + `d/dx` (x2).
= (2 × 3) x(3-1) - (3 × 4) x(4-1) - 2x(2-1)
= 6 x2 - 12 x3 + 2 x
`d/dx` (2x3 - 3x4 + x2) = 6 x2 - 12 x3 + 2 x
Answer: The derivative of given quadratic function is `d/dx` (2x3 - 3x4 + x2) = 6x2 - 12 x3 + 2 x
Derivatives of quadratic function problem 2:
Find the derivative of quadratic function y = 3t2 + t3 - 30t. with respect to 't'.Is this topic free algebra help hard for you? Watch out for my coming posts.
Solution:
Given Quadratic function is y = 3t2 + t3 - 30t.
dy = (3 × 2)t(2 - 1) dt + (1 × 3)t(3 - 1) dt - 30t(1 - 1) dt.
= 6t dt + 3t2 dt - 30dt.
= (6t + 3t2 - 30) dt
`dy/(dt)` = (6t + 3t2 - 30)
Answer: The derivative of given quadratic function is `d/dt` (3 t2 + t3 - 30t) = 3t2 + 6t - 30.
Derivatives of quadratic function problem 3:
Determine the second derivative of given quadratic function f(x) = 15x2 + 10x + 5 with respect to x.
Solution:
Given quadratic function is, f(x) = 15x2 + 10x + 5
The First derivative of quadratic function is f'
f' =` (df )/ (dx)` = (15 × 2)x(2 - 1) + (10 × 1)x (1-1)
= 30x + 10
The Second derivative of quadratic function is, f''
` d^2/dx` f(x) = `d^2/dx` ( 15x2 + 10x + 5)
= `d/dx` (f')
= `d/dx` ( 30x + 10)
= ` d/dx` (30x) +` d/dx` (10)
= 30 `d/dx` (x) + `d/dx` (10)
= 30 + 0
= 30
Answer: The Second derivative of given quadratic function is 30
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