## Related questions with answers

Calculate $\frac{d y}{d x}$. You need not expand your answers. $y=\frac{3 x-1}{(x-5)(x-4)(x-1)}$

Solution

VerifiedRecall the $\textbf{quotient rule}$:

$\dv{x}\qty(\dfrac{f(x)}{g(x)}) = \dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$

Note that we are going to have to differentiate the function in the denominator. Let's do it separately to make the solution more 'clean'.

We will use the $\textbf{product rule}$ for 3 functions:$\dv{x}(f(x)g(x)h(x)) = f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$

We now get

$\begin{align*} \qty[(x-5)(x-4)(x-1)]' &= (x-5)'(x-4)(x-1)+(x-5)(x-4)'(x-1)+(x-5)(x-4)(x-1)'\\ &=(x-4)(x-1)+(x-5)(x-1)+(x-5)(x-4)\\ &=(x^2-5x+4)+(x^2-6x+5)+(x^2-9x+20)\\ &=3x^2-20x+29 \end{align*}$

Now, let's compute the original derivative using quotient rule.

$\begin{align*} \dv{y}{x} &= \qty[\dfrac{3x-1}{(x-5)(x-4)(x-1)}]'\\ &=\dfrac{(3x-1)'(x-5)(x-4)(x-1)+(3x-1)[(x-5)(x-4)(x-1)]'}{[(x-5)(x-4)(x-1)]^2}\\ &=\dfrac{3(x-5)(x-4)(x-1)+(3x-1)(3x^2-20x+29)}{(x-5)^2(x-4)^2(x-1)^2} \end{align*}$

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