If $y=\sqrt{\dfrac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}}$, find $\dfrac{dy}{dx}$.

$\dfrac12\,y\left[\dfrac{1}{x-3}+\dfrac{2x}{x^{2}+4}-\dfrac{6x+4}{3x^{2}+4x+5}\right]$

CBSE Class 12 · 2006 · DifferentiationLong Answer

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

, find

\frac{d y}{d x}

Answer: $\frac{1}{2} y [\frac{1}{x - 3} + \frac{2 x}{x ^{2} + 4} - \frac{6 x + 4}{3 x ^{2} + 4 x + 5}]$

Take logarithm and use

lo g = \frac{1}{2} lo g

to split the product/quotient:

lo g y = \frac{1}{2} [lo g (x - 3) + lo g (x^{2} + 4) - lo g (3 x^{2} + 4 x + 5)]

Differentiate w.r.t.

x

\frac{1}{y} \frac{d y}{d x} = \frac{1}{2} [\frac{1}{x - 3} + \frac{2 x}{x ^{2} + 4} - \frac{6 x + 4}{3 x ^{2} + 4 x + 5}]

Multiply by

y

\frac{d y}{d x} = \frac{1}{2} \frac{( x - 3 ) ( x ^{2} + 4 )}{3 x ^{2} + 4 x + 5} [\frac{1}{x - 3} + \frac{2 x}{x ^{2} + 4} - \frac{6 x + 4}{3 x ^{2} + 4 x + 5}]

CBSE Class 12 Board · 2006 · Subjective