Question

Differentiate $\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$ w.r.t.$x$

Answer 1

Ley $y=\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$

Taking log⁡ on both sides, we get,

$\log y=\log \sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$

$\log y=\log {\left(\frac{(x-3)(x^2+4)}{3x^2+4x+5}\right)}^{\frac{1}{2}}$

$\log y=\frac{1}{2}\log \frac{(x-3)(x^2+4)}{(3x^2+4x+5)}$

$(\text{Using}\log a^n=n\log a)$

$\log y=\frac{1}{2}\log \frac{(x-3)(x^2+4)}{(3x^2+4x+5)}$

$\left(\begin{array}{c}\text{Using}\log ab=\log a+\log b\\ \text{and}\log \frac{a}{b}=\log a-\log b\end{array}\right)$

$\log y=\frac{1}{2}\left(\begin{array}{c}\log (x-3)+\log (x^2+4)-\\ \log (3x^2+4x+5)\end{array}\right)\cdots \left(i\right)$

Differentiating $\left(i\right)$ w.r.t $x$, we get,

$\frac{d\left(\log y\right)}{dx}=\frac{1}{2}\left(\begin{array}{c}\frac{d\left(\log \right(x-3\left)\right)}{dx}+\frac{d\left(\log \right(x^2+4\left)\right)}{dx}\\ -\frac{d\left(\log \right(3x^2+4x+5\left)\right)}{dx}\end{array}\right)$

$(\text{using chain rule}\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx})$

$\frac{1}{y}\left(\frac{dy}{dx}\right)=\frac{1}{2}\left(\begin{array}{c}\frac{1}{x-3}\cdot \frac{d(x-3)}{dx}+\frac{1}{x^2+4}\cdot \frac{d(x^2+4)}{dx}\\ -\frac{1}{(3x^2+4x+5)}\cdot \frac{d(3x^2+4x+5)}{dx}\end{array}\right)$

$\frac{1}{y}\left(\frac{dy}{dx}\right)=\frac{1}{2}\left[\begin{array}{c}\frac{1}{x-3}(1-0)+\frac{1}{x^2+4}(2x+0)\\ -\frac{1}{(3x^2+4x+5)}(6x+4+0)\end{array}\right]$

$\frac{1}{y}\left(\frac{dy}{dx}\right)=\frac{1}{2}(\frac{1}{x-3}+\frac{2x}{x^2+4}-\frac{6x+4}{3x^2+4x+5})$

$\frac{dy}{dx}=\frac{y}{2}(\frac{1}{(x-3)}+\frac{2x}{x^2+4}-\frac{6x+4}{3x^2+4x+5})$

Substituting the value of $y=\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}$,

We get,

$\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}\times (\frac{1}{(x-3)}+\frac{2x}{x^2+4}-\frac{6x+4}{3x^2+4x+5})$