use the definition of derivative to find d/dx(1/(3-x))
at x+h
y = 1/(3-x-h)
at x
y = 1/(3-x)
subtract
1/(3-x-h) -1/(3-x)
[(3-x) -(3-x-h)]/[(3-x-h)(3-x)]
[ h]/[ x^2-6x+xh-3h+9]
divide by h
1/[x^2-6x+9 +xh-3h]
let h --->0
1/(x-3)^2
the dividing by h is confusing me. wouldn't you be left with (x^2/h)-(6x/h)+x-3+(9/h)?
To find the derivative of the function 1/(3-x), we can use the definition of derivative. The derivative of a function f(x) at a point x is defined as the limit of the difference quotient as h approaches zero:
f'(x) = lim(h->0) [f(x+h) - f(x)] / h
Let's apply this definition to find the derivative of 1/(3-x):
f(x) = 1/(3-x)
First, let's find f(x+h):
f(x+h) = 1/(3-(x+h))
Next, let's calculate the difference quotient:
[f(x+h) - f(x)] / h = [1/(3-(x+h)) - 1/(3-x)] / h
The next step involves simplifying the difference quotient. To do that, we need a common denominator:
[f(x+h) - f(x)] / h = [(3-x) - (3-(x+h))] / [(3-(x+h))(3-x)]
Simplifying further:
[f(x+h) - f(x)] / h = [(3-x) - (3-x-h)] / [(3-(x+h))(3-x)]
Now let's combine like terms:
[f(x+h) - f(x)] / h = [h] / [(3-(x+h))(3-x)]
Finally, let's take the limit of this difference quotient as h approaches zero:
f'(x) = lim(h->0) [h] / [(3-(x+h))(3-x)]
To evaluate this limit, we can substitute h=0:
f'(x) = [0] / [(3-x)(3-x)]
Simplifying further:
f'(x) = 0 / [(3-x)(3-x)]
Therefore, the derivative of 1/(3-x) is 0.