Question

Find the derivatives w.r.t.x:
$\rm \frac{ax^{2}+bx+c}{px^{2}+qx+r}$

Answer 1

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Answer 2

Answer:

$\rm \dfrac{d}{dx}\left ( \dfrac{ax^{2}+bx+c}{px^{2}+qx+r}\right)=\dfrac{(aq-bp)x^2+2(ar-pc)x+(br-qc)}{(px^{2}+qx+r)^2}.$

Step-by-step explanation:

Given function:

$\rm \dfrac{ax^{2}+bx+c}{px^{2}+qx+r}$

We know that the derivative of a function which is in fraction, such as, $\rm \dfrac{f(x)}{g(x)}$ is given by

$\rm \dfrac{d}{dx}\left ( \dfrac {f(x)}{g(x)}\right) = \dfrac{g(x)\dfrac{d}{dx}(f(x))-f(x)\dfrac{d}{dx}(g(x))}{g^2(x)}$.

Here, we have,

$\rm f(x) = ax^{2}+bx+c\\g(x)=px^{2}+qx+r}$

Therefore,

$\rm \dfrac{d}{dx}\left ( \dfrac{ax^{2}+bx+c}{px^{2}+qx+r}\right) = \dfrac{(px^{2}+qx+r)\dfrac{d}{dx}(ax^{2}+bx+c)-(ax^{2}+bx+c)\dfrac{d}{dx}(px^{2}+qx+r)}{(px^{2}+qx+r)^2}\\=\dfrac{(px^{2}+qx+r)(2ax+b)-(ax^{2}+bx+c)(2px+q)}{(px^{2}+qx+r)^2}\\=\dfrac{(2apx^{3}+2aqx^2+2arx+bpx^{2}+bqx+br)-(2pax^{3}+2pbx^2+2pcx+qax^{2}+qbx+qc)}{(px^{2}+qx+r)^2}\\=\dfrac{(aqx^2+2arx-bpx^{2}+br-2pcx-qc)}{(px^{2}+qx+r)^2}\\=\dfrac{(aq-bp)x^2+2(ar-pc)x+(br-qc)}{(px^{2}+qx+r)^2}.$