### Mathematics Class 10 Lab Manual | 21 Lab Activities

Mathematics Lab Manual Class X   lab activities for class 10 with complete observation Tables strictly according to the CBSE syllabus also very useful & helpful for the students and teachers.

## Math Assignment  Class XI  Ch - 12

Extra questions of chapter 12 Limits & Derivatives class 11  with answers and  hints to the difficult questions, strictly according to the CBSE and DAV syllabus. Important and useful math. assignment for the students of class 11

## Strictly according to the CBSE and DAV Board

Question 1:  Differentiate the following  by first principal

(i)  sin 2x    (ii)  x ex    (iii)  x2ex    (iv) Cos2x     (v) x2cosx

Question 2 : Using first principle find the derivative of  x sinx

Ans: x cosx + sinx

Question 3 : Find the derivative of sin2x w.r.t. x, by first principle.

Ans: 2sinxcosx = sin2x

Question 4 : Using first principle find the derivative of sin (x2 + 1).

Ans: 2x cos (x2 + 1)

Question 5 : Find the derivative of cos(5x + 2) w.r.t x by first principle.

Ans: - 5sin(5x + 2

Question 6 : Find the derivative of tan(2x + 3) w.r.t. x, by first principle

Ans: 2sec2(2x + 3)

Question 7 : Find the derivative of tan (ax + b) from the first principle.

Ans: asec2(ax + b)

Question 8 : Find the derivative of f(x) = $\sqrt{sinx}$  from first principle.

Ans: $\frac{cosx}{2\sqrt{sinx}}$

Question 9 : Find the derivative of f(x) = $\sqrt{cosx}$  from first principle.

Ans:  $\frac{-sinx}{2\sqrt{cosx}}$

Question 10 : Find the derivative of $f(x)= \sqrt{tanx}$  using first principal
Ans:   $\large \dpi{110}\frac{sec^{2}x}{2\sqrt{tanx}}$
Question 11 : Find the derivative of  f(x) = $sin\sqrt{x}$  with respect to x using first principal.

Ans:  $\frac{cos\sqrt{x}}{2\sqrt{x}}$

Solution hint for this question is given at the end.

Question 12 : Differentiate : $y=\frac{(x^{3}+1)(x-2)}{x^{2}}$

Ans :  $2x-2-\frac{1}{x^{2}}+\frac{4}{x^{3}}$

Question 13 : Differentiate: $y=e^{xloga}+e^{alogx}+e^{aloga}$

Ans :    aLoga + axa-1.

Question 14 : Differentiate: $y=\left ( sin\frac{x}{2}+cos\frac{x}{2} \right )^{2}$

Ans : cosx

Question 15 : Differentiate:  $y=\frac{x^{3}sinx}{cosx}$

Ans : 3x2tanx + x3sec2x

Question 16 : Differentiate: $y=x^{2}\; sinx \; logx\]$

Ans : 2x sinx logx + x2 cosx logx + x sinx

Question 17 : Differentiate:  $y=x^{5}\; e^{x} +x^{6}\; logx$

Ans : x4ex (5 + x) + x5(6logx + 1)

Question 18 : Differentiate: $y=\frac{1+tanx}{1-tanx}$
Ans : $\frac{2sec^{2}x}{(1-tanx)^{2}}$
Question 19 : Differentiate: $y=\frac{x+e^{x}}{1+logx}$
Ans:

$\bg{black}\frac{xlogx(1+e^{x})+e^{x}(x-1)}{x(1+logx)^{2}}$

Question 20 : Differentiate:  $\frac{sin(x+a)}{cosx}$

Ans: cosa.sec2x

Question 21 : If y = xsinx + cosx then find  $\frac{dy}{dx}$  at  x = $\frac{\pi }{2}$

Ans: 0

Question 22 : Find the derivative of  cosx - sinx at x = $\frac{2\pi }{3}$

Ans: $-\frac{\sqrt{3}}{2}+\frac{1}{2}$

Question 23 : Find the derivative of  $\frac{x-1}{x^{2}}$  w.r.t x

Ans: $-\frac{1}{x^{2}}+\frac{2}{x^{3}}$

Question 24 : Find the derivative of f (x) = $\frac{x+2}{x}$ w. r. t. x

Ans: $-\frac{2}{x^{2}}$

Question 25 : Differentiate: $y=(x+secx)(x-tanx)\]$

Question 26 : Differentiate: $y=(1-2tanx)(5+4sinx)\]$

Question 27 : Differentiate:  $y=\frac{e^{x}-tanx}{cotx-x^{n}}$

Question 28 : Differentiate : y =  $\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}$  with respect to x

Question 29 : If   $y=\frac{sin(x+9^{o})}{cosx}$ , then find $\frac{dy}{dx}$  at x = 0
Question 30 : If  $y=x+\frac{1}{x}$ , then prove that  $x^{2}\frac{dy}{dx}-xy+2=0$

Question 31 :
If f(x) = kx2 -5x +4 and f’(3) = 37, find the value of k.

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SOLUTION HINT

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Formula to find the derivative by using first principal

$f'(x)=\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

Product Rule to find the derivative

$\frac{d}{dx}(uv)=\frac{du}{dx}v+u\frac{dv}{dx}$

Quotient rule to find the derivative

$\frac{d}{dx}\left ( \frac{u}{v} \right )=\frac{\frac{du}{dx}v-u\frac{dv}{dx}}{v^{2}}$

Solution Hint Q11

$f(x)=sin\sqrt{x}$

$f(x+h)=sin\sqrt{x+h}$

Now by definition of first principle, we have

$f'(x)=\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

$f(x)'=\displaystyle \lim_{x \to 0}\frac{sin\sqrt{x+h}-sin\sqrt{x}}{h}$

$f(x)'=\displaystyle \lim_{x \to 0}\frac{1}{h}\left [ sin\sqrt{x+h}-sin\sqrt{x} \right ]$

Applying  CD formula here, we get

$f(x)'=\displaystyle \lim_{x \to 0}\frac{1}{h}\left [ 2cos\left ( \frac{\sqrt{x+h}+\sqrt{x}}{2} \right )sin\left ( \frac{\sqrt{x+h}-\sqrt{x}}{2} \right ) \right ]$

Putting the limit in the first cos function we get

$f(x)'=2cos \left ( \frac{2\sqrt{x}}{2} \right )\displaystyle \lim_{x \to 0}\frac{1}{h}\left [ sin\left ( \frac{\sqrt{x+h}-\sqrt{x}}{2} \right ) \right ]$

Rationalizing the numerator of the angle of sine function, we get

$f(x)'=2cos \left ( \sqrt{x} \right )\displaystyle \lim_{x \to 0}\frac{1}{h}\left [ sin\left ( \frac{\sqrt{x+h}-\sqrt{x}}{2}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \right ) \right ]$

Simplify the above equation we get

$f(x)'=2cos \left ( \sqrt{x} \right )\displaystyle \lim_{x \to 0}\frac{1}{h}\left [ sin\left ( \frac{x+h-x}{2(\sqrt{x+h}+\sqrt{x})} \right ) \right ]$

$f(x)'=cos \left ( \sqrt{x} \right )\displaystyle \lim_{x \to 0}\left [ \frac{sin\left ( \frac{h/2}{\sqrt{x+h}+\sqrt{x}} \right )}{h/2} \right ]$
Now making the denominator similar to the angle of the sine function, we get

$f(x)'=cos \left ( \sqrt{x} \right )\displaystyle \lim_{x \to 0}\left [ \frac{sin\left ( \frac{h/2}{\sqrt{x+h}+\sqrt{x}} \right )}{h/2\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}} \right ]$

$f(x)'=cos \left ( \sqrt{x} \right )\displaystyle \lim_{x \to 0}\left [ \frac{sin\left ( \frac{h/2}{\sqrt{x+h}+\sqrt{x}} \right )}{ \frac{h/2}{\sqrt{x+h}+\sqrt{x}}} \right ]\times \frac{1}{\sqrt{x+h}+\sqrt{x}}$

Taking limit and applying the formula $\displaystyle \lim_{ x\to 0}\frac{sinx}{x}=1$ , we get
$f(x)'=cos \sqrt{x}\times 1\times \frac{1}{\sqrt{x}+\sqrt{x}}$

$f(x)'=\frac{cos \sqrt{x}}{2\sqrt{x}}$