Math Assignment Class XI Ch -13

Math Assignment Class XI Ch - 13
DERIVATIVES

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

MATHEMATICS ASSIGNMENT ON

DERIVATIVES (XI)

Strictly according to the CBSE and DAV Board

Question 1: Differentiate the following by first principal

(i) sin 2x (ii) x e^x (iii) x²e^x (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: $\frac{sec^{2}x}{2\sqrt{tanx}}$

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 : a^xLoga + ax^n-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 : x⁴e^x (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:

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

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

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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

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

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

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