Differentiate the following function from first principles
f(x)= 5x/(3-x)
just plug and chug
f(x+h) - f(x) = 5(x+h)/(3-x-h) - 5x/(3-x)
= (5(x+h)(3-x) - 5x(3-x-h)) / (3-x-h)(3-x)
= 15h / (3-x-h)(3-h)
Now divide by h and you have
15 / (3-x-h)(3-x)
Now take the limit as h->0 and you have
15/(3-x)^2
To differentiate the function f(x) = (5x)/(3-x) from first principles, we need to find the derivative of the function using:
lim(h→0) [f(x+h) - f(x)] / h
Let's start by finding f(x+h):
f(x+h) = 5(x+h) / (3 - (x+h))
= 5(x+h) / (3 - x - h)
Now, we can find f(x+h) - f(x):
f(x+h) - f(x) = [5(x+h) / (3 - x - h)] - [5x / (3 - x)]
= 5x + 5h + 15 - 3x - 5h / (x - 3)(x + h - 3)
= 2x + 15 / (x - 3)(x + h - 3)
Next, we can substitute f(x+h) - f(x) into the limit expression:
lim(h→0) [f(x+h) - f(x)] / h
= lim(h→0) [2x + 15 / (x - 3)(x + h - 3)] / h
To simplify, let's first deal with the denominator:
(x - 3)(x + h - 3) = x^2 + hx - 3x - 3h - 3hx + 9
= x^2 + hx - 3x - 3h - 3hx + 9
Now, we can substitute this expression back into the limit:
lim(h→0) [2x + 15 / x^2 + hx - 3x - 3h - 3hx + 9] / h
Expanding the numerator:
lim(h→0) [2x + 15] / [x^2 + hx - 3x - 3h - 3hx + 9] / h
We can now simplify the denominator by factoring:
(x^2 + hx - 3x - 3h - 3hx + 9) = (x^2 + 2hx - 6x - 3h + 9)
Now, we substitute this back into the limit:
lim(h→0) [2x + 15] / [(x^2 + 2hx - 6x - 3h + 9)] / h
Now, we can cancel out the common h factor in the denominator:
lim(h→0) [2x + 15] / [(x^2 + 2hx - 6x - 3h + 9)]
Taking the limit as h approaches 0, we can substitute h = 0:
[2x + 15] / [(x^2 - 6x + 9)]
= 2x + 15 / x^2 - 6x + 9
Therefore, the derivative of f(x) = (5x) / (3 - x) from first principles is:
f'(x) = 2x + 15 / x^2 - 6x + 9