NCERT Exemplar Solutions Class 11 Maths Limits and Derivatives — Short and Long Answer Type Q29–46

NCERT Exemplar Problems, Class 11 Mathematics — Limits and Derivatives, Short and Long Answer Type questions 29 to 46. All 18 questions solved.

The first fourteen questions (29 to 42) are routine differentiation: the sum, product and quotient rules and the standard derivatives of $x^n$, $\sin x$, $\cos x$, $\tan x$, $\cot x$, $\sec x$. The last four (43 to 46) are the same kind of function but must be differentiated from first principle, meaning from the definition $f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$, with no shortcut rules allowed.

The printed key has a typographical slip in Question 34 (“sec$^4$” where $x^4$ is meant); the answer itself is as worked below.

Key insight. Before differentiating, simplify. Question 29 is a polynomial plus $\frac1x$ once the division is carried out, Question 32 collapses to $\tan^2x$, Question 38 to $1+\sin2x$ and Question 41 to $\frac18\sin^32x$. Fewer terms means fewer chances for a sign slip. For first-principle questions, the whole job is to rewrite the difference $f(x+h)-f(x)$ so that a factor $h$ appears and cancels.

Differentiate each of the functions with respect to $x$ in Questions 29 to 42.

Question 29

$\dfrac{x^4+x^3+x^2+1}{x}$

Solution. Divide first: $\dfrac{x^4+x^3+x^2+1}{x}=x^3+x^2+x+x^{-1}$. Now differentiate term by term.

$$\frac{d}{dx}\left(x^3+x^2+x+x^{-1}\right)=3x^2+2x+1-\frac1{x^2}$$

Question 30

$\left(x+\dfrac1x\right)^3$

Solution. Expanding gives $x^3+3x+\dfrac3x+\dfrac1{x^3}$, because $(p+q)^3=p^3+3p^2q+3pq^2+q^3$ with $pq=1$. Differentiate each term:

$$3x^2+3-\frac3{x^2}-\frac3{x^4}$$

(The chain rule gives the same: $3\left(x+\frac1x\right)^2\left(1-\frac1{x^2}\right)$.)

Question 31

$(3x+5)(1+\tan x)$

Solution. Product rule: $(uv)’=u’v+uv’$ with $u=3x+5$, $v=1+\tan x$, $u’=3$, $v’=\sec^2x$.

$$3(1+\tan x)+(3x+5)\sec^2x=3x\sec^2x+5\sec^2x+3\tan x+3$$

Question 32

$(\sec x-1)(\sec x+1)$

Solution. This is a difference of squares: $\sec^2x-1=\tan^2x$. Differentiating $\tan^2x$ by the chain rule gives $2\tan x\sec^2x$.

Question 33

$\dfrac{3x+4}{5x^2-7x+9}$

Solution. Quotient rule: $\left(\dfrac uv\right)’=\dfrac{u’v-uv’}{v^2}$ with $u=3x+4$, $v=5x^2-7x+9$, $v’=10x-7$.

$$\frac{3(5x^2-7x+9)-(3x+4)(10x-7)}{(5x^2-7x+9)^2}=\frac{15x^2-21x+27-30x^2-19x+28}{(5x^2-7x+9)^2}=\frac{55-40x-15x^2}{(5x^2-7x+9)^2}$$

Question 34

$\dfrac{x^5-\cos x}{\sin x}$

Solution. Quotient rule with $u=x^5-\cos x$, $u’=5x^4+\sin x$, $v=\sin x$, $v’=\cos x$:

$$\frac{(5x^4+\sin x)\sin x-(x^5-\cos x)\cos x}{\sin^2x}=\frac{5x^4\sin x-x^5\cos x+\sin^2x+\cos^2x}{\sin^2x}$$

and $\sin^2x+\cos^2x=1$. The printed key shows “$5\sec^4x\sin x$” in the numerator; that is a typographical slip for $5x^4\sin x$, since no $\sec x$ can arise from this function.

Question 35

$\dfrac{x^2\cos\frac\pi4}{\sin x}$

Solution. $\cos\frac\pi4=\frac1{\sqrt2}$ is a constant, not a function of $x$, so it comes out of the derivative. Then the quotient rule on $\dfrac{x^2}{\sin x}$:

$$\frac1{\sqrt2}\cdot\frac{2x\sin x-x^2\cos x}{\sin^2x}=\frac{x}{\sqrt2}\cdot\frac{1}{\sin x}\left(2-x\cot x\right)=\frac{x}{\sqrt2}\operatorname{cosec}x\,(2-x\cot x)$$

Question 36

$(ax^2+\cot x)(p+q\cos x)$

Solution. Product rule, with $a,p,q$ constants: $u=ax^2+\cot x$, $u’=2ax-\operatorname{cosec}^2x$; $v=p+q\cos x$, $v’=-q\sin x$.

$$(ax^2+\cot x)(-q\sin x)+(p+q\cos x)(2ax-\operatorname{cosec}^2x)$$

Question 37

$\dfrac{a+b\sin x}{c+d\cos x}$

Solution. Quotient rule: $u=a+b\sin x$, $u’=b\cos x$; $v=c+d\cos x$, $v’=-d\sin x$.

$$\frac{b\cos x(c+d\cos x)+d\sin x(a+b\sin x)}{(c+d\cos x)^2}=\frac{bc\cos x+bd\cos^2x+ad\sin x+bd\sin^2x}{(c+d\cos x)^2}$$

and $bd\cos^2x+bd\sin^2x=bd$.

Question 38

$(\sin x+\cos x)^2$

Solution. Differentiate by the chain rule: $2(\sin x+\cos x)(\cos x-\sin x)=2(\cos^2x-\sin^2x)=2\cos2x$. (Expanding first, $(\sin x+\cos x)^2=1+\sin2x$, whose derivative is the same $2\cos2x$.)

Question 39

$(2x-7)^2(3x+5)^3$

Solution. Product rule with the chain rule for each factor:

$$2(2x-7)\cdot2\,(3x+5)^3+(2x-7)^2\cdot3(3x+5)^2\cdot3$$

Take out the common factor $(2x-7)(3x+5)^2$:

$$(2x-7)(3x+5)^2\left[4(3x+5)+9(2x-7)\right]=(2x-7)(3x+5)^2(30x-43)$$

Question 40

$x^2\sin x+\cos2x$

Solution. Product rule on the first term and the chain rule on the second: $\dfrac{d}{dx}\cos2x=-2\sin2x$.

$$x^2\cos x+2x\sin x-2\sin2x$$

Question 41

$\sin^3x\cos^3x$

Solution. Since $\sin x\cos x=\frac12\sin2x$, the function is $\dfrac18\sin^32x$. By the chain rule:

$$\frac18\cdot3\sin^22x\cdot2\cos2x=\frac34\sin^22x\cos2x$$

Question 42

$\dfrac1{ax^2+bx+c}$

Solution. Write it as $(ax^2+bx+c)^{-1}$ and apply the chain rule: $-(ax^2+bx+c)^{-2}\cdot(2ax+b)$.

Differentiate each of the functions with respect to $x$ in Questions 43 to 46 using first principle.

Question 43

$\cos(x^2+1)$

Solution. Here $f(x+h)-f(x)=\cos\left((x+h)^2+1\right)-\cos(x^2+1)$. Use $\cos A-\cos B=-2\sin\frac{A+B}2\sin\frac{A-B}2$ with $A=(x+h)^2+1$ and $B=x^2+1$. Then $A-B=h(2x+h)$ and $\frac{A+B}2=x^2+xh+\frac{h^2}2+1$.

$$f'(x)=\lim_{h\to0}\frac{-2\sin\left(x^2+xh+\frac{h^2}2+1\right)\sin\frac{h(2x+h)}2}{h}$$

Write $\dfrac{\sin\frac{h(2x+h)}2}{h}=\dfrac{\sin\frac{h(2x+h)}2}{\frac{h(2x+h)}2}\cdot\dfrac{2x+h}2$. As $h\to0$ the first factor tends to $1$ and the second to $x$, and the other sine tends to $\sin(x^2+1)$.

$$f'(x)=-2\sin(x^2+1)\cdot x=-2x\sin(x^2+1)$$

Question 44

$\dfrac{ax+b}{cx+d}$

Solution. Put the two fractions over one denominator:

$$f(x+h)-f(x)=\frac{(ax+ah+b)(cx+d)-(ax+b)(cx+ch+d)}{(cx+ch+d)(cx+d)}.$$

In the numerator the terms not involving $h$ cancel: $(ax+b)(cx+d)-(ax+b)(cx+d)=0$. What is left is $ah(cx+d)-(ax+b)ch=h\,(ad-bc)$. So

$$f'(x)=\lim_{h\to0}\frac{h(ad-bc)}{h\,(cx+ch+d)(cx+d)}=\frac{ad-bc}{(cx+d)^2}.$$

Question 45

$x^{2/3}$

Solution. The fractional power is awkward, so substitute cube roots: let $u=(x+h)^{1/3}$ and $v=x^{1/3}$, so that $u\to v$ as $h\to0$, $(x+h)^{2/3}=u^2$ and $h=u^3-v^3$.

$$f'(x)=\lim_{u\to v}\frac{u^2-v^2}{u^3-v^3}=\lim_{u\to v}\frac{u+v}{u^2+uv+v^2}=\frac{2v}{3v^2}=\frac2{3v}=\frac23x^{-1/3}$$

Question 46

$x\cos x$

Solution. Split the difference by adding and subtracting $x\cos(x+h)$:

$$\frac{(x+h)\cos(x+h)-x\cos x}{h}=\frac{x\left[\cos(x+h)-\cos x\right]}{h}+\cos(x+h).$$

The bracket divided by $h$ is the first-principle form of $\frac{d}{dx}\cos x=-\sin x$ (via $\cos(x+h)-\cos x=-2\sin\left(x+\frac h2\right)\sin\frac h2$ and $\frac{\sin(h/2)}{h/2}\to1$). As $h\to0$, $\cos(x+h)\to\cos x$.

$$f'(x)=-x\sin x+\cos x$$

Common mistakes

• Question 33 and 34. Getting the quotient rule’s numerator in the wrong order. It is $u’v-uv’$, and swapping the two terms reverses every sign.
• Question 35. Differentiating $\cos\frac\pi4$ as if it were a function of $x$, and writing $-\sin\frac\pi4$. It is the constant $\frac1{\sqrt2}$.
• Question 36 and 37. Differentiating $a,b,c,d,p,q$ as if they were variables. Only $x$ varies.
• Question 39. Leaving the answer as a long unsimplified sum. Both terms carry $(2x-7)$ and $(3x+5)^2$; taking those out gives the tidy $(30x-43)$ factor and an answer that can be checked against the key.
• Question 43. Writing $\cos\left((x+h)^2+1\right)=\cos(x^2+1)+\cos h^2$. A cosine of a sum does not split into a sum; use the product formula for $\cos A-\cos B$.
• Question 45. Expanding $(x+h)^{2/3}$ as a binomial. That series is not available at this stage; the substitution with cube roots is what makes the difference factorise.

Practise next

• Other pages in this unit: Short Answer Type Q1–14, Short Answer Type Q15–28, Long Answer and Objective Type Q47–64 and Objective and Fill in the Blanks Q65–80.
• NCERT textbook: Exercise 12.2 (derivatives, including first-principle questions) and the Miscellaneous Exercise.