SN Dey Class 11 Solutions Differentiation Very Short Answer Type Questions

SN Dey Class 11 Differentiation Solutions | Solution of Class 11 Derivative | SN Dey Class 11 Solutions | SN Dey Class 11 Derivative Solutions Very Short Answer Type Questions | West Bengal Board Class 11 Derivative Solution | SN Dey Class 11 Math Chapter Derivative Solution | Differentiation Very Short Answer Type Questions Answers | sn dey class 11 derivatives solutions

Click here to get SN Dey Class 11 All Chapters Solutions

SN Dey Class 11 Differentiation Very Short Answer Type Questions Solutions

SN Dey class 11 Differentiation very short answer type questions Ex 1 solutions:

Ex 1: Define the derivative of a function f(x) at an arbitrary point x in its domain of definition.

Solution:

Let x be a point in the domain of the definition of f(x). The derivative of f(x) at a point x is given by the following limit:

$\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$

and it is denoted by $f'(x)$.

SN Dey class 11 Differentiation very short answer type questions Ex 2 solutions:

Ex 2: Examine whether f(x)=|x| has a derivative at x=0.

Solution:

Let f(x)=|x|. We know that |x|=x if x>0 and |x|=-x if x $\leq$ 0. The left hand derivative of f(x) at x=0 is equal to

Left derivative of |x| at x=0

And the right hand derivative of f(x) at x=0 is equal to

Right derivative of |x| at x=0

As the left hand derivative and the right hand derivative of f(x)=|x| at x=0 are not equal, we conclude that f(x)=|x| has no derivative at x=0. In other words,

$f'(0)$ does not exist.

SN Dey class 11 Differentiation very short answer type questions Ex 3 solutions:

Ex 3: If $y=\frac{1}{x}$, find $\frac{dy}{dx}$; also find the value of x for which dy/dx becomes undefined.

Solution:

derivative of y=1/x

SN Dey class 11 Differentiation very short answer type questions Ex 4 solutions:

sn dey-11-differentiation-solution-vsatq-Ex 4

SN Dey class 11 Differentiation very short answer type questions Ex 5 solutions:

sn dey-11-differentiation-solution-vsatq-Ex 5

Ex 6: Differentiate the following functions w.r.t. x:

Ex 6(i): If y=x16 then find dy/dx.

Solution:

dy/dx= d/dx(x16)

= 16 x16-1 by the power rule of derivatives.

= 16 x15

Ex 6(ii): If y=3x7 then find dy/dx.

Solution:

dy/dx= d/dx(3x7)

= 3 ⋅ 7x7-1 by the power rule of derivatives.

= 21 x6

 

sn dey-11-differentiation-vsatq-Ex 6.(iii)

 

sn dey-11-differentiation-vsatq-Ex 6.(iii)

 

sn dey-11-differentiation-vsatq-Ex 6.(v)

 

sn dey-11-differentiation-vsatq-Ex 6.(vi)

 

Ex 6(vii): If y=(x2-2)2 then find dy/dx.

Solution:

dy/dx= d/dx{(x2-2)2}

= 2(x2 -2) ⋅ d/dx(x2-2)

= 2(x2 -2) ⋅ (2x-0)

= 4x(x2 -2) 

 

sn dey-11-differentiation-vsatq-Ex 6.(viii)

 

Ex 6(ix): If y=(x2-2x)(x+1) then find dy/dx.

Solution:

dy/dx= d/dx{(x2-2x)(x+1)}

= (x2 -2x) d/dx(x+1) + (x+1) d/dx(x2 -2x), by the product rule of derivatives.

= (x2 -2x)(1+0) + (x+1)(2x-2)

= x2 -2x + 2x2 +2x-2x-2

= 3x2 -2x – 2

 

sn dey-11-differentiation-vsatq-Ex 6.(x)

 

sn dey-11-differentiation-vsatq-Ex 6.(xi)

 

sn dey-11-differentiation-vsatq-Ex 6.(xii)

 

Ex 7: Find dy/dx :

Ex 7(i): If y=3log x -4√x+2ex then find dy/dx.

Solution:

y = 3log x -4√x+2ex

∴ dy/dx= d/dx(3log x -4√x+2ex)

= 3 d/dx(log x) – 4 d/dx(√x)+2 d/dx(ex)

= $3 \cdot \frac{1}{x}-$ 4 d/dx(x1/2) + 2ex

= $\frac{3}{x} – 4 \cdot \frac{1}{2}x^{1/2-1}$ + 2ex

= $\frac{3}{x} -2x^{-1/2}$ + 2ex

= $\frac{3}{x} -2x^{-1/2}$ + 2ex

Ex 7(ii): If y = log2 x  then find dy/dx.

Solution:

y = log2

∴ dy/dx= d/dx(log2 x)

= d/dx(log2 e ⋅ loge x)  [∵ logab = logae ⋅logeb]

= log2 e ⋅ d/dx(loge x)

= log2 e ⋅ 1/x =1/x log2e

Ex 7(iii): If y = 2xm-3mx+4ex  then find dy/dx.

Solution:

y = 2xm-3mx+4ex

∴ dy/dx= d/dx(2xm-3mx+4ex)

= 2 ⋅ d/dx(xm) – 3⋅ d/dx(mx)+ 4⋅ d/dx(ex)  

= 2 ⋅ mxm-1 – 3⋅ mx logem + 4⋅ ex

= 2mxm-1 – 3mx logem + 4ex

Ex 7(iv): If y = 2x+2 -ex+1 +3log 2x  then find dy/dx.

Solution:

y = 2x+2 -ex+1 +3log 2x

∴ dy/dx= d/dx(2x+2 -ex+1 +3log 2x)

= d/dx(2x+2) – d/dx(ex+1)+ 3⋅ d/dx(log 2x)  

= 2x+2 ⋅ loge2 ⋅ d/dx(x+2) – ex+1 ⋅d/dx(x+1) + $3\cdot \frac{1}{2x}\cdot$ d/dx(2x) 

= 2x+2 loge2 ⋅ 1 – ex+1 ⋅1 + $3\cdot \frac{1}{2x}\cdot$ 2 

= 2x+2 loge2 – ex+1 + $\frac{3}{x}$ 

Ex 7(v): If y = log10 x +10x +x10+10  then find dy/dx.

Solution:

y = log10 x +10x +x10+10

∴ dy/dx= d/dx(log10 x +10x +x10+10)

= d/dx(log10 x) +d/dx(10x) + d/dx(x10) +d/dx(10)

= $ \frac{1}{x}$ log10e + 10xloge10 + 10 x10-1 + 0

= $ \frac{1}{x}$ log10e + 10xloge10 + 10x9 

Ex 7(vi): If y = ex+1 -5x+1 +elog x+logax +log xa  then find dy/dx.

Solution:

y = ex+1 -5x+1 +elog x+logax +log xa

∴ dy/dx= d/dx(ex+1 -5x+1 +elog x+logax +log xa)

= d/dx(ex+1) -d/dx(5x+1) + d/dx(x) +d/dx(logax) +d/dx(log xa[∵ elog x = x]

= ex+1 d/dx(x+1) – 5x+1 loge5 ⋅ d/dx(x+1) + 1 + $\frac{1}{x}$ logae +axa-1 d/dx(xa

= ex+1(1+0) – 5x+1 loge5 ⋅(1+0) + 1 + $\frac{1}{x}$ logae + $\frac{1}{x^a}$ axa-1

= ex+1 – 5x+1 loge5 + 1 + $\frac{1}{x}$ logae + $\frac{a}{x}$

Ex 7(vii): If y = a sec x + b tan x – c cosec x + d cot x -e  then find dy/dx.

Solution:

y = a sec x + b tan x – c cosec x + d cot x -e

∴ dy/dx= d/dx(a sec x + b tan x – c cosec x + d cot x -e)

= d/dx(a sec x) + d/dx(b tan x) – d/dx(c cosec x) + d/dx(d cot x) – d/dx(e)

= a d/dx(sec x) + b d/dx(tan x) -c d/dx(cosec x) +d d/dx(cot x) – 0

= a sec x tan x + b sec2x +c cosec x cot x -d cosec2x

 

sn dey-11-differentiation-vsatq-Ex 7.(viii)

 

sn dey-11-differentiation-vsatq-Ex 7.(ix)

 

sn dey-11-differentiation-vsatq-Ex 7.(x)

 

sn dey-11-differentiation-vsatq-Ex 7.(xi)

SN Dey class 11 Differentiation very short answer type questions Ex 8 solutions:

Ex 8: Find the derivative of cot x by expressing it in the form cos x ⋅ cosec x.

Solution:

sn dey-11-differentiation-vsatq-Ex 8 solution

SN Dey class 11 Differentiation very short answer type questions Ex 9 solutions:

sn dey-11-differentiation-vsatq-Ex 9 solution

SN Dey class 11 Differentiation very short answer type questions Ex 10 solutions:

sn-dey-11-differentiation-vsatq-Ex-10 solution

SN Dey class 11 Differentiation very short answer type questions Ex 11 solutions:

sn-dey-11-differentiation-vsatq-Ex-11 solution

SN Dey class 11 Differentiation very short answer type questions Ex 12 solutions:

sn-dey-11-differentiation-vsatq-Ex-12 solution

SN Dey class 11 Differentiation very short answer type questions Ex 13 solutions:

sn-dey-11-differentiation-vsatq-Ex-13 solution

SN Dey class 11 Differentiation very short answer type questions Ex 14 solutions:

Ex 14: If $y=x^5$, show that $x \frac{dy}{dx}-5y=0$.

Solution:

y=x^5 then x dy/dx-5y=0

 

sn dey-11-differentiation-solution-vsatq-Ex 15.(i)

 

sn dey-11-differentiation-solution-vsatq-Ex 15.(ii)

 

sn dey-11-differentiation-solution-vsatq-Ex 15.(iii)

 

sn dey-11-differentiation-vsatq-Ex 15.(iv)

Solution:

Given f(x)=mx+c

∴ f(0) = m⋅0+c =c

So f(0)=1 implies that c=1

∴ f(x) = mx+1

Differentiating f(x) with respect to x, we get that

$f'(x) = m$

Now $f'(0)=1$ implies that m=1

Thus, f(x)=x+1

∴ f(2) = 2+1 =3

Share via:
WhatsApp Group Join Now
Telegram Group Join Now