If $y=\sqrt{x^{2}+1}-\log\!\left(\dfrac{1}{x}+\sqrt{1+\dfrac{1}{x^{2}}}\right)$, find $\dfrac{dy}{dx}$.

Differentiation — CBSE Class 12 · 2008 Board Question with Solution

y = x^{2} + 1 - lo g (\frac{1}{x} + 1 + \frac{1}{x ^{2}})

, find

\frac{d y}{d x}

Answer: $\frac{x ^{2} + 1}{x}$

Simplify the logarithmic term first:

\frac{1}{x} + 1 + \frac{1}{x ^{2}} = \frac{1 + x ^{2} + 1}{x}

, so

y = x^{2} + 1 - lo g (1 + x^{2} + 1) + lo g x

Differentiate term by term (chain rule on the surd and the log):

\frac{d y}{d x} = \frac{x}{x ^{2} + 1} - \frac{1}{1 + x ^{2} + 1} \cdot \frac{x}{x ^{2} + 1} + \frac{1}{x}

Combine the first two terms over

x^{2} + 1 (1 + x^{2} + 1)

and simplify the numerator to

x x^{2} + 1

\frac{d y}{d x} = \frac{x}{x ^{2} + 1} \cdot \frac{x ^{2} + 1}{1 + x ^{2} + 1} + \frac{1}{x} = \frac{x}{1 + x ^{2} + 1} + \frac{1}{x}

Putting over a common denominator and using

(x^{2} + 1)^{2} - 1 = x^{2}

gives the compact result:

\frac{d y}{d x} = \frac{x ^{2} + 1}{x}

CBSE Class 12 Board · 2008 · Subjective