#### Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 44 maths

$\frac{1}{a \sin ^{2} t \cos t}$

Hint:

You must know the derivative of sin, cos, tan and logarithm function

Given:

\begin{aligned} &x=a \sin t \\ &y=a\left(\cos t+\log \tan \frac{t}{2}\right) \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}

Solution:

\begin{aligned} &x=a \sin t \\ &\frac{d x}{d t}=a \cos t \\ &y=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &\frac{d y}{d t}=a\left(-\sin t+\cot \frac{t}{2} \times \sec ^{2} \frac{t}{2} \times \frac{t}{2}\right) \end{aligned}

\begin{aligned} &=a\left(-\sin t+\frac{1}{2 \sin \left(\frac{t}{2}\right) \times \cos \left(\frac{t}{2}\right)}\right) \\ &\frac{d y}{d t}=a\left(-\sin t+\frac{1}{\sin t}\right) \\ &=a\left(\frac{-\sin ^{2} t+1}{\sin t}\right) \Rightarrow \frac{a \cos ^{2} t}{\sin t} \end{aligned}

\begin{aligned} &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{\frac{a \cos ^{2} t}{\sin t}}{a \cos t} \\ &=\frac{\cos t}{\sin t}=\cot t \\ &\text { Again } \frac{d^{2} y}{d x^{2}}=-cosec^{2} t \frac{d t}{d x}=cose c^{2} t \times \frac{1}{a \cos t} \\ &=\frac{1}{a \sin ^{2} t \cos t} \end{aligned}