### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

## Math Assignment  Class XI  Ch - 13DERIVATIVES

Extra questions of chapter 13 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:  $\frac{cos\sqrt{x}}{2\sqrt{x}}$

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 + axn-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

******************************************

SOLUTION HINT

*****************************************

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
$f(x)'=cos \sqrt{x}\times 1\times \frac{1}{\sqrt{x}+\sqrt{x}}$

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