How do you find the derivative of #x=((s-1)(s-2))/(s-3)#?

Question

Christopher Allers · Accepted Answer

#(dx)/(ds)=(s^2-6s+7)/(s-3)^2# According to Quotient rule if #f(x)=(g(x))/(h(x))# then #(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2# Here we have #x=((s-1)(s-2))/(s-3)=(s^2-3s+2)/(s-3)# Hence #(dx)/(ds)=((2s-3)(s-3)-(s^2-3s+2)xx1)/(s-3)^2# = #(2s^2-9s+9-s^2+3s-2)/(s-3)^2# = #(s^2-6s+7)/(s-3)^2#

Charles Arocha · Answer

undefined To find the derivative of $x = \frac{{(s - 1)(s - 2)}}{{s - 3}}$, you can use the quotient rule. The quotient rule states that if you have a function of the form $f(x) = \frac{{g(x)}}{{h(x)}}$, then its derivative is given by $$f'(x) = \frac{{g'(x)h(x) - g(x)h'(x)}}{{[h(x)]^2}}.$$

Applying the quotient rule to the given function $x = \frac{{(s - 1)(s - 2)}}{{s - 3}}$, where $g(s) = (s - 1)(s - 2)$ and $h(s) = s - 3$, we get:

$$x' = \frac{{(s - 1)(s - 2)'(s - 3) - (s - 1)(s - 2)(s - 3)'}}{{(s - 3)^2}}.$$

Now, differentiate $g(s) = (s - 1)(s - 2)$ and $h(s) = s - 3$, which gives:

$$(s - 1)(s - 2)' = (1)(s - 2) + (s - 1)(1) = s - 2 + s - 1 = 2s - 3,$$

and $$(s - 3)' = 1.$$

Substitute these derivatives back into the quotient rule formula:

$$x' = \frac{{(2s - 3)(s - 3) - (s - 1)(s - 2)}}{{(s - 3)^2}}.$$

Now, expand and simplify:

$$x' = \frac{{2s^2 - 6s - 3s + 9 - (s^2 - 3s - 2s + 6)}}{{(s - 3)^2}}.$$

$$x' = \frac{{2s^2 - 9s + 9 - s^2 + 5s - 6}}{{(s - 3)^2}}.$$

$$x' = \frac{{s^2 - 4s + 3}}{{(s - 3)^2}}.$$

So, the derivative of $x = \frac{{(s - 1)(s - 2)}}{{s - 3}}$ with respect to $s$ is $x' = \frac{{s^2 - 4s + 3}}{{(s - 3)^2}}$.