Get Answers to all your Questions

header-bg qa

Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 5 maths textbook solution

Answers (1)


        2cos\: x\: cosec^{3}x


You must know about how third derivatives be find


If  y=log x(sin x). Prove that

        \frac{d^{3}y}{dx^{3}}=2cos\: x\: cosec^{3}x


Let\: \: y=log (sin x)\; \; \; \; \; \; \; \; (\frac{d\: log\: x}{dx}=\frac{1}{x})

        \begin{aligned} &\frac{d y}{d x}=\frac{1}{\sin x} \frac{d}{d x} \sin x \\ &\frac{d y}{d x}=\frac{1}{\sin x} \cos x \\ &\frac{d y}{d x}=\cot x \end{aligned}

again differentiating we get

        \frac{d^{2} y}{d x^{2}}=-cosec^{2} x                        (\frac{d\: cot\: x}{dx}=-cosec^{2}x)

Again differentiating w.r.t. x we get

        \begin{aligned} &\frac{d^{3} y}{d x^{3}}=-2 \operatorname{cosec} x\cdot(-\cos ec x\cdot \cot x) \quad\left(\frac{d \operatorname{cosec}^{2} x}{d x}=-cosec\: x \cot x\right) \\ &\frac{d^{3} y}{d x^{3}}=-2 \cos x \operatorname{cosec}^{3} x \end{aligned}

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support