To find the derivative of a function using first principles, we need to find the limit of the difference quotient as h approaches 0.
The difference quotient for a function f(x) is given by:
[f(x + h) - f(x)] / h
Let's apply this formula to the given function:
f(x) = -8x² - 5x + 13
[f(x + h) - f(x)] / h
= [(-(8(x + h)²) - 5(x + h) + 13) - ( -8x² - 5x + 13)] / h
= [(-(8(x² + 2hx + h²)) - 5x - 5h + 13) - (-8x² - 5x + 13)] / h
= [(-8x² - 16hx - 8h² - 5x - 5h + 13) - (-8x² - 5x + 13)] / h
= [-8x² - 16hx - 8h² - 5x - 5h + 13 + 8x² + 5x - 13] / h
= (-16hx - 8h² - 5h) / h
Now, we can take the limit as h approaches 0:
lim(h->0) (-16hx - 8h² - 5h) / h
= lim(h->0) -16x - 8h - 5
= -16x - 5
Therefore, the derivative of -8x² - 5x + 13 using first principles is -16x - 5.
Find the derivative of -8x²-5x+13 using first principles
5 answers
Derivative of 1/1-yx using first principles
To find the derivative of the function f(x) = 1/(1 - yx) using first principles, we need to find the limit of the difference quotient as h approaches 0.
The difference quotient for a function f(x) is given by:
[f(x + h) - f(x)] / h
Let's apply this formula to the given function:
f(x) = 1/(1 - yx)
[f(x + h) - f(x)] / h
= [(1/(1 - y(x + h))) - (1/(1 - yx))] / h
To simplify further, we need to find a common denominator:
= [((1 - yx) - (1 - y(x + h)))/[(1 - yx)(1 - y(x + h))]] / h
= [(1 - yx - 1 + y(x + h))/[(1 - yx)(1 - y(x + h))]] / h
= [(y(h))/[(1 - yx)(1 - y(x + h))]] / h
Now, we can take the limit as h approaches 0:
lim(h->0) [(y(h))/[(1 - yx)(1 - y(x + h))]] / h
= lim(h->0) [(y(h))/[(1 - yx)(1 - yx - yh)]] / h
= lim(h->0) (y(h))/[(1 - yx)(1 - yx)] / h
= y / (1 - yx)^2
Therefore, the derivative of the function 1/(1 - yx) using first principles is y / (1 - yx)^2.
The difference quotient for a function f(x) is given by:
[f(x + h) - f(x)] / h
Let's apply this formula to the given function:
f(x) = 1/(1 - yx)
[f(x + h) - f(x)] / h
= [(1/(1 - y(x + h))) - (1/(1 - yx))] / h
To simplify further, we need to find a common denominator:
= [((1 - yx) - (1 - y(x + h)))/[(1 - yx)(1 - y(x + h))]] / h
= [(1 - yx - 1 + y(x + h))/[(1 - yx)(1 - y(x + h))]] / h
= [(y(h))/[(1 - yx)(1 - y(x + h))]] / h
Now, we can take the limit as h approaches 0:
lim(h->0) [(y(h))/[(1 - yx)(1 - y(x + h))]] / h
= lim(h->0) [(y(h))/[(1 - yx)(1 - yx - yh)]] / h
= lim(h->0) (y(h))/[(1 - yx)(1 - yx)] / h
= y / (1 - yx)^2
Therefore, the derivative of the function 1/(1 - yx) using first principles is y / (1 - yx)^2.
Derivative of √x using first Principles
To find the derivative of the function f(x) = √x using first principles, we need to find the limit of the difference quotient as h approaches 0.
The difference quotient for a function f(x) is given by:
[f(x + h) - f(x)] / h
Let's apply this formula to the given function:
f(x) = √x
[f(x + h) - f(x)] / h
= [√(x + h) - √x] / h
To simplify further, we can use the difference of squares formula:
= [(√(x + h) - √x)(√(x + h) + √x)] / (h * (√(x + h) + √x))
Using the formula (a - b)(a + b) = a^2 - b^2, we can simplify the numerator:
= [(x + h) - x] / (h * (√(x + h) + √x))
= [h] / (h * (√(x + h) + √x))
= 1 / (√(x + h) + √x)
Now, we can take the limit as h approaches 0:
lim(h->0) 1 / (√(x + h) + √x)
= 1 / (2√x)
Therefore, the derivative of √x using first principles is 1 / (2√x).
The difference quotient for a function f(x) is given by:
[f(x + h) - f(x)] / h
Let's apply this formula to the given function:
f(x) = √x
[f(x + h) - f(x)] / h
= [√(x + h) - √x] / h
To simplify further, we can use the difference of squares formula:
= [(√(x + h) - √x)(√(x + h) + √x)] / (h * (√(x + h) + √x))
Using the formula (a - b)(a + b) = a^2 - b^2, we can simplify the numerator:
= [(x + h) - x] / (h * (√(x + h) + √x))
= [h] / (h * (√(x + h) + √x))
= 1 / (√(x + h) + √x)
Now, we can take the limit as h approaches 0:
lim(h->0) 1 / (√(x + h) + √x)
= 1 / (2√x)
Therefore, the derivative of √x using first principles is 1 / (2√x).
1 answer