Question

Calculate dy/dx.You need not expand your answer.
y = x/(x − 7)(x − 4)

Answer 1

not sure what all is in the denominator

Answer 2

\[\frac{x}{(x-7)(x-4)} ?\]

Answer 3

thats what i kindof thought he meant

Answer 4

use quotient rule is one way

Answer 5

Its dealing with derivatives I'm not sure what to use and how to solve it.

Answer 6

you have f/g
so use quotient rule

(f'g-g'f)/(g^2)

Answer 7

I will show you three ways!
----------------------
1st way: Quotient Rule!

\[y=\frac{x}{(x-7)(x-4)}=\frac{x}{x^2-11x+28}\]

\[y'=\frac{(x)'(x^2-11x+28)-x(x^2-11x+28)'}{(x^2-11x+28)^2}\]
\[=\frac{1(x^2-11x+28)-x(2x-11+0)}{(x^2-11x+28)^2}\]
------------------------
2nd way: Product Rule!
\[y=x(x^2-11x+28)^{-1}\]
\[y'=(x)'(x^2-11x+28)^{-1}+x[(x^2-11x+28)^{-1}]'\]
\[y'=1(x^2-11x+28)^{-1}+x(-1)(x^2-11x+28)^{-1-1}(x^2-11x+28)'\]
\[=\frac{1}{x^2-11x+28}-x \frac{1}{(x^2-11x+28)^2}(2x-11+0)\]
------------------------
3rd way: Logarithmic Differentiation (my favorite!)
\[y=\frac{x}{(x-7)(x-4)}\]
Before Differentiating both sides we will do ln( ) of both sides
\[\ln(y)=\ln(\frac{x}{(x-7)(x-4)})\]
\[\ln(y)=\ln(x)-\ln[(x-7)(x-4)]\]
\[\ln(y)=\ln(x)-[\ln(x-7)+\ln(x-4)]\]
\[\ln(y)=\ln(x)-\ln(x-7)-\ln(x-4)\]
Now the derivative part!
\[\frac{y'}{y}=\frac{1}{x}-\frac{1}{x-7}-\frac{1}{x-4}\]
Now we wanted y' so to isolate y' multiply y on both sides
\[y'=y(\frac{1}{x}-\frac{1}{x-7}-\frac{1}{x-4})\]
But remember \[y=\frac{x}{(x-7)(x-4)}\]

\[y'=\frac{x}{(x-7)(x-4)} (\frac{1}{x}-\frac{1}{x-7}-\frac{1}{x-4})\]