If $y=\log(1+\cos x)$, prove that $\dfrac{d^{3}y}{dx^{3}}+\dfrac{d^{2}y}{dx^{2}}\cdot\dfrac{dy}{dx}=0$.

CBSE Class 12 · 2005 · DifferentiationLong Answer

y = lo g (1 + cos x)

, prove that

\frac{d ^{3} y}{d x ^{3}} + \frac{d ^{2} y}{d x ^{2}} \cdot \frac{d y}{d x} = 0

Answer: Proved

Differentiate once:

\frac{d y}{d x} = \frac{- sin x}{1 + cos x} = - tan \frac{x}{2}

(using

sin x = 2 sin \frac{x}{2} cos \frac{x}{2}

and

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

).
Second derivative:

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

Third derivative:

\frac{d ^{3} y}{d x ^{3}} = - \frac{1}{2} \cdot 2 sec \frac{x}{2} \cdot sec \frac{x}{2} tan \frac{x}{2} \cdot \frac{1}{2} = - \frac{1}{2} sec^{2} \frac{x}{2} tan \frac{x}{2}

Now form the required combination:

\frac{d ^{3} y}{d x ^{3}} + \frac{d ^{2} y}{d x ^{2}} \cdot \frac{d y}{d x} = - \frac{1}{2} sec^{2} \frac{x}{2} tan \frac{x}{2} + (- \frac{1}{2} sec^{2} \frac{x}{2}) (- tan \frac{x}{2}) = 0

\frac{d ^{3} y}{d x ^{3}} + \frac{d ^{2} y}{d x ^{2}} \cdot \frac{d y}{d x} = 0

CBSE Class 12 Board · 2005 · Subjective