RD Sharma Solutions Class 12 Mathematics Chapter 11 FBQ

Edited By Satyajeet Kumar | Updated on Jan 27, 2022 02:54 PM IST

The RD sharma solutions are well researched and tested textbooks for the thorough preparations for CBSE class 12 students. Class 12 RD Sharma chapter 11 exercise FBQ solution, deals with the chapter of Higher-order derivatives which is one of the intricate chaptemrs for the students to follow. The RD Sharma class 12th exercise FBQ comes quite handy with the solutions of your complex problems while solving math chapter 11.RD Sharma Solutions So, it is advised to go through the RD Sharma class 12 solution of Higher-Order derivatives exercise FBQ for clearing out the concepts and have a better understanding about the chapter.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

RD Sharma Class 12 Solutions Chapter 11 FBQ Higher Order Derivatives - Other Exercise

Higher Order Derivatives Excercise: FBQ

Higher Order Derivatives Exercise Fill in the blanks Question 1

Answer: $\frac{d^{2} y}{d x^{2}}=\frac{5}{16 t^{6}}$
Hint: Differentiate x & y w.r.t t. Use $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$ . Again, differentiate w.r.t x.
Given: $y=t^{10}+1 \& x=t^{8}+1$
Solution:
It is given that: $y=t^{10}+1 \& x=t^{8}+1$
Differentiating y w.r.t t,
$\frac{d y}{d t}=10 t^{9}$ …(i) $\left[\because \frac{d}{d x} p^{n}=n p^{n-1}\right]$
Differentiating x w.r.t t,
$\frac{d x}{d t}=8 t^{7}$ …(ii)
So, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$
$\begin{aligned} &\frac{d y}{d x}=\frac{10 t^{9}}{8 t^{7}} \\ & \end{aligned}$
$\frac{d y}{d x}=\frac{5}{4} t^{2}$ …(iii)
Again, differentiate (iii) w.r.t x:
$\begin{aligned} &\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{5}{4} t^{2}\right) \\ & \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=\left(\frac{5}{4}\right) \frac{d}{d t}\left(t^{2}\right) \cdot \frac{d t}{d x}$
$\begin{aligned} &=\left(\frac{5}{4}\right)(2 t) \cdot \frac{1}{8 t^{7}} \\ & \end{aligned}$ $\left[\because \frac{d x}{d t}=8 t^{7}\right]$
$=\frac{10}{32 t^{6}}$
Thus, $\frac{d^{2} y}{d x^{2}}=\frac{5}{16 t^{6}}$

Higher Order Derivatives Exercise Fill in the blanks Question 2

Answer: $\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}} \sec ^{3} \theta$
Hint: Differentiate x & y w.r.t $\theta$ . Use $\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$ . Again, differentiate w.r.t x.
Given: $y=b \cos \theta \& \quad x=a \sin \theta$
Solution:
It is given that: $x=a \sin \theta \& y=b \cos \theta$
Differentiating y w.r.t t,
$\frac{d y}{d \theta}=-b \sin \theta$ …(i)
Differentiating y w.r.t t,
$\frac{d x}{d \theta}=a \cos \theta$ …(ii)
So, $\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$
$\frac{d y}{d x}=\frac{-b \sin \theta}{a \cos \theta}$ …(iii)
Again, differentiate (iii) w.r.t x:
$\begin{aligned} &\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{-b \sin \theta}{a \cos \theta}\right) \\ & \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=\frac{d}{d \theta}\left(\frac{-b \sin \theta}{a \cos \theta}\right) \cdot \frac{d \theta}{d x}$
$\begin{aligned} &=\frac{d}{d \theta}\left(\frac{-b}{a} \tan \theta\right) \cdot \frac{1}{a \cos \theta} \\ & \end{aligned}$ [ using (ii) ]
$=\frac{-b}{a} \sec ^{2} \theta \cdot\left(\frac{1}{a \cos \theta}\right) \\$
$=\frac{-b}{a^{2}} \sec ^{3} \theta$
Thus, $\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}} \sec ^{3} \theta$

Higher Order Derivatives Exercise Fill in the blanks Question 3

Answer: $\frac{d^{2} y}{d x^{2}}=e^{x}$
Hint: Differentiate y w.r.t x twice to get $\frac{d^{2} y}{d x^{2}}$ .
Given: $y=x+e^{x}$
Solution:
It is given that: $y=x+e^{x}$
$\therefore \frac{d y}{d x}=1+e^{x}$
Again, differentiating w.r.t x:
$\therefore \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(1+e^{x}\right)$
Thus, $\frac{d^{2} y}{d x^{2}}=e^{x}$ .

Higher Order Derivatives exercise Fill in the blanks question 4

Answer: $\frac{d^{2} y}{d x^{2}}=e^{-x}$
Hint: Differentiate y w.r.t x twice to get $\frac{d^{2} y}{d x^{2}}$
Given: $y=1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots \ldots$
Solution:
It is given that: $y=1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots \ldots$
$\therefore \frac{d y}{d x}=-1+\frac{2 x}{2 !}-\frac{3 x^{2}}{3 !}+\frac{4 x^{3}}{4 !}-\ldots$
Again, differentiating w.r.t x:
$\begin{aligned} &\therefore \frac{d^{2} y}{d x^{2}}=\frac{2}{2 !}-\frac{3(2) x}{3 !}+\frac{4(3) x^{2}}{4 !}-\ldots \\ & \end{aligned}$
$=\frac{2}{2}-\frac{6 x}{6}+\frac{12 x^{2}}{24}-\ldots \\$
$=1-x+\frac{x^{2}}{2 !}-\ldots .$
We know that:
$e^{-x}=1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots$
Thus, $\frac{d^{2} y}{d x^{2}}=e^{-x}$ .

Higher Order Derivatives exercise Fill in the blanks question 5

Answer: $\frac{d^{2} y}{d x^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}}$
Hint: Differentiate given equation w.r.t x. Use $\frac{d x}{d y}=\frac{1}{d y / d x}$
Given: $y=x+e^{x}$
Solution:
It is given that: $y=x+e^{x}$
$\begin{aligned} &\therefore \frac{d y}{d x}=1+e^{x} \\ & \end{aligned}$
$\frac{d x}{d y}=\frac{1}{d y / d x} \\$
$\therefore \frac{d x}{d y}=\frac{1}{1+e^{x}}$ …(i)
Again, differentiating w.r.t x:
$\begin{aligned} &\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d}{d y}\left(\frac{1}{1+e^{x}}\right) \\ & \end{aligned}$
$\frac{d^{2} x}{d y^{2}}=\frac{d}{d x}\left(\frac{1}{1+e^{x}}\right) \frac{d x}{d y} \\$
$=\left(\frac{-1}{\left(1+e^{x}\right)^{2}}\right) \frac{d}{d x}\left(e^{x}\right) \cdot\left(\frac{1}{1+e^{x}}\right)$ [ using (i) ]
$=\left(\frac{-1}{\left(1+e^{x}\right)^{2}}\right)\left(e^{x}\right) \cdot\left(\frac{1}{1+e^{x}}\right)$
Thus, $\frac{d^{2} x}{d y^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}}$

Higher Order Derivatives exercise Fill in the blanks question 6

Answer: $\frac{d^{2} y}{d x^{2}}=\frac{-1}{x^{2}}$
Hint: Differentiate y w.r.t x twice to get $\frac{d^{2} y}{d x^{2}}$ .
Given: $y=\log _{e} x$
Solution:
It is given that: $y=\log _{e} x$
$\therefore \frac{d y}{d x}=\frac{1}{x}$
Again, differentiating w.r.t x:
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-1}{x^{2}} \\ & \end{aligned}$
Thus, $2 \frac{d^{2} y}{d x^{2}}=\frac{-1}{x^{2}} .$ .

The RD Sharma class 12th exercise FBQ is the fill in the blank question type that provides very few questions but the important ones that might be asked in the exams. The RD Sharma class 12th exercise FBQ

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. Is the class 12 RD Sharma chapter 11 FBQ solution easily accessible by everyone?

The presence of the RD Sharma books on the Careers360 website has made it freely accessible for everyone. Also, it can be downloaded from the same website

2. Can the students who are preparing for JEE mains have a touch with the RD Sharma class 12 chapter 11 ex FBQ?

The students who are preparing for the JEE mains and other entrance exams can have a good amount of practice using the sums given in the RD Sharma books.

3. How often does the RD Sharma class 12 chapter 11 solution be updated?

The RD Sharma class FBQ solution is updated each time the syllabus or questions of the NCERT book gets changed.

4. What role does the Class 12 Chapter 11 FBQ material play in making the students complete their homework?

This set of books can be used to recheck the answers that the students arrive at. It also helps them in gaining an idea of how a problem should be solved.

5. How can the students use the class 12 RD Sharma chapter 11 FBQ in the best way?

The students can use the class 12 RD Sharma chapter 11 FBQ solution book to practice several questions apart from those given in the textbooks.