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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=x ^{16} then find dy/dx.**

Solution:

dy/dx= d/dx(x^{16})

= 16 x^{16-1} by the power rule of derivatives.

= 16 x^{15}

**Ex 6(ii): If y=3x ^{7} then find dy/dx.**

Solution:

dy/dx= d/dx(3x^{7})

= 3 ⋅ 7x^{7-1} by the power rule of derivatives.

= 21 x^{6}

**Ex 6(vii): If y=(x ^{2}-2)^{2} then find dy/dx.**

Solution:

dy/dx= d/dx{(x^{2}-2)^{2}}

= 2(x^{2} -2) ⋅ d/dx(x^{2}-2)

= 2(x^{2} -2) ⋅ (2x-0)

= 4x(x^{2} -2)

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

Solution:

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

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

= (x^{2} -2x)(1+0) + (x+1)(2x-2)

= x^{2} -2x + 2x^{2} +2x-2x-2

= 3x^{2} -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 = log _{2} x then find dy/dx.**

Solution:

y = log_{2} x

∴ dy/dx= d/dx(log_{2} x)

= d/dx(log_{2} e ⋅ log_{e} x) [∵ log_{a}b = log_{a}e ⋅log_{e}b]

= log_{2} e ⋅ d/dx(log_{e} x)

= log_{2} e ⋅ 1/x =1/x log_{2}e

**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_{e}m + 4⋅ e^{x}

= 2mx^{m-1} – 3m^{x }log_{e}m + 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_{e}2 ⋅ d/dx(x+2) – e^{x+1} ⋅d/dx(x+1) + $3\cdot \frac{1}{2x}\cdot$ d/dx(2x)

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

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

**Ex 7(v): If y = log _{10} x +10^{x }+x^{10}+10 then find dy/dx.**

Solution:

y = log_{10} x +10^{x }+x^{10}+10

∴ dy/dx= d/dx(log_{10} x +10^{x }+x^{10}+10)

= d/dx(log_{10} x) +d/dx(10^{x}) + d/dx(x^{10}) +d/dx(10)

= $ \frac{1}{x}$ log_{10}e + 10^{x}log_{e}10 + 10 x^{10-1} + 0

= $ \frac{1}{x}$ log_{10}e + 10^{x}log_{e}10 + 10x^{9}

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

Solution:

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

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

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

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

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

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

**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 sec^{2}x +c cosec x cot x -d cosec^{2}x

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

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.