SN Dey Class 11 Solutions Differentiation Very Short Answer Type Questions

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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

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

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:

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

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

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 x^16-1 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 ⋅ 7x^7-1 by the power rule of derivatives.

= 21 x⁶

 

 

 

 

 

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) 

 

 

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

 

 

 

 

Ex 7: Find dy/dx :

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

Solution:

y = 3log x -4√x+2e^x

∴ dy/dx= d/dx(3log x -4√x+2e^x)

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

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

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

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

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

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

Solution:

y = log₂ x 

∴ dy/dx= d/dx(log₂ x)

= d/dx(log₂ e ⋅ log_e x)  [∵ log_ab = log_ae ⋅log_eb]

= log₂ e ⋅ d/dx(log_e x)

= log₂ e ⋅ 1/x =1/x log₂e

Ex 7(iii): If y = 2x^m-3m^x+4e^x  then find dy/dx.

Solution:

y = 2x^m-3m^x+4e^x

∴ dy/dx= d/dx(2x^m-3m^x+4e^x)

= 2 ⋅ d/dx(x^m) – 3⋅ d/dx(m^x)+ 4⋅ d/dx(e^x)  

= 2 ⋅ mx^m-1 – 3⋅ m^x log_em + 4⋅ e^x

= 2mx^m-1 – 3m^x log_em + 4e^x

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

Solution:

y = 2^x+2 -e^x+1 +3log 2x

∴ dy/dx= d/dx(2^x+2 -e^x+1 +3log 2x)

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

= 2^x+2 ⋅ log_e2 ⋅ d/dx(x+2) – e^x+1 ⋅d/dx(x+1) + $3\cdot \frac{1}{2x}\cdot$ d/dx(2x) 

= 2^x+2 log_e2 ⋅ 1 – e^x+1 ⋅1 + $3\cdot \frac{1}{2x}\cdot$ 2 

= 2^x+2 log_e2 – e^x+1 + $\frac{3}{x}$ 

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

Solution:

y = log₁₀ x +10^x +x¹⁰+10

∴ dy/dx= d/dx(log₁₀ x +10^x +x¹⁰+10)

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

= $ \frac{1}{x}$ log₁₀e + 10^xlog_e10 + 10 x^10-1 + 0

= $ \frac{1}{x}$ log₁₀e + 10^xlog_e10 + 10x⁹ 

Ex 7(vi): If y = e^x+1 -5^x+1 +e^{log x}+log_ax +log x^a  then find dy/dx.

Solution:

y = e^x+1 -5^x+1 +e^{log x}+log_ax +log x^a

∴ dy/dx= d/dx(e^x+1 -5^x+1 +e^{log x}+log_ax +log x^a)

= d/dx(e^x+1) -d/dx(5^x+1) + d/dx(x) +d/dx(log_ax) +d/dx(log x^a)  [∵ e^{log x} = x]

= e^x+1 d/dx(x+1) – 5^x+1 log_e5 ⋅ d/dx(x+1) + 1 + $\frac{1}{x}$ log_ae +ax^a-1 d/dx(x^a) 

= e^x+1(1+0) – 5^x+1 log_e5 ⋅(1+0) + 1 + $\frac{1}{x}$ log_ae + $\frac{1}{x^a}$ ax^a-1

= e^x+1 – 5^x+1 log_e5 + 1 + $\frac{1}{x}$ log_ae + $\frac{a}{x}$

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

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

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

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

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

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:

 

 

 

 

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