Question

Calculate dydx\frac{d y}{d x}. You need not expand your answers. y=3x1(x5)(x4)(x1)y=\frac{3 x-1}{(x-5)(x-4)(x-1)}

Solution

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Answered 2 years ago
Answered 2 years ago
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Recall the quotient rule\textbf{quotient rule}:

d ⁣dx(f(x)g(x))=f(x)g(x)f(x)g(x)[g(x)]2\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 product rule\textbf{product rule} for 3 functions:d ⁣dx(f(x)g(x)h(x))=f(x)g(x)h(x)+f(x)g(x)h(x)+f(x)g(x)h(x)\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

[(x5)(x4)(x1)]=(x5)(x4)(x1)+(x5)(x4)(x1)+(x5)(x4)(x1)=(x4)(x1)+(x5)(x1)+(x5)(x4)=(x25x+4)+(x26x+5)+(x29x+20)=3x220x+29\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.

 ⁣dy ⁣dx=[3x1(x5)(x4)(x1)]=(3x1)(x5)(x4)(x1)+(3x1)[(x5)(x4)(x1)][(x5)(x4)(x1)]2=3(x5)(x4)(x1)+(3x1)(3x220x+29)(x5)2(x4)2(x1)2\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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