RD Sharma Solutions Class 12 Mathematics Chapter 11 FBQ

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 Solutions Class 12 Mathematics Chapter 11 FBQ

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

Higher Order Derivatives Excercise: FBQ

Confused between CGPA and Percentage?

RD Sharma Chapter-wise Solutions

Related eBooks and Sample Papers

Applications for Admissions are open.

Aakash Re-NEET 2026 Batch

University of Liverpool, Bengaluru Campus

Victoria University, Delhi NCR

Illinois Tech Mumbai

University of Aberdeen Mumbai

University of York, Mumbai

News and Notifications

Latest updates

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

NCERT Syllabus

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

CISCE 12th

Schools By Medium

By Ownership

By City

By State

IMO

IEO

NSO

NMMS

NTSE

NVS

JNVST

Download Careers360 App

All this at the convenience of your phone

Scan and download the app