Class 11 RD Sharma Solutions- Chapter 30 Derivatives - Exercise 30.1

Question 1. Find the derivative of f(x) = 3x at x = 2

Solution:

Given: f(x)=3x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h   {where h is a small positive number}

Derivative of f(x)=3x at x=2 is given as:

f'(2)= \lim_{h \to 0} \frac {f(2+h)-f(2)} h

⇒ f'(2)= \lim_{h \to 0} \frac {3(2+h)-3(2)} h

⇒ f'(2)= \lim_{h \to 0} \frac {3h+6-6} h

⇒ f'(2)= \lim_{h \to 0} \frac {3h} h

⇒ f'(2)= \lim_{h \to 0} 3 = 3

Hence, derivative of f(x)=3x at x=2 is 3

Question 2. Find the derivative of f(x) = x²– 2 at x = 10

Solution:

Given: f(x)= x²-2 

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h   {where h is a small positive number}

Derivative of f(x)=x²-2 at x=10 is given as:

f'(10)= \lim_{h \to 0} \frac {f(10+h)-f(10)} h

⇒f'(10)= \lim_{h \to 0} \frac {(100+h^2+20h-2-100+2))} h

⇒f'(10)= \lim_{h \to 0} \frac {h^2+20h} h

⇒f'(10)= \lim_{h \to 0} \frac {h(h+20)} h

⇒f'(10)= \lim_{h \to 0} {h+20}

⇒f'(10)= 0+20=20

Hence, derivative of f(x)=x²-2 at x=10 is 20

Question 3. Find the derivative of f(x) = 99x at x = 100

Solution:

Given: f(x)= 99x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h   {where h is a small positive number}

Derivative of f(x)=99x at x=100 is given as:

f'(100)= \lim_{h \to 0} \frac {f(100+h)-f(100)} h

⇒ f'(100)= \lim_{h \to 0} \frac {99(100+h)-99(100)} h

⇒ f'(100)= \lim_{h \to 0} \frac {9900+99h-9900} h

⇒ f'(100)= \lim_{h \to 0} 99 = 99

Hence, derivative of f(x)=99x at x=100 is 99

Question 4. Find the derivative of f(x) = x at x = 1

Solution:

Given: f(x)=x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h  {where h is a small positive number}

Derivative of f(x)=x at x=1 is given as:

f'(1)= \lim_{h \to 0} \frac {f(1+h)-f(1)} h

⇒ f'(1)= \lim_{h \to 0} \frac {(1+h)-1} h

⇒ f'(1)= \lim_{h \to 0} \frac {h} h

⇒f'(1)= \lim_{h \to 0} 1 = 1

Hence, derivative of f(x)=x at x=1 is 1

Question 5. Find the derivative of f(x) = \cos x  at x = 0

Solution:

Given: f(x)=\cos x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h  {where h is a small positive number}

Derivative of f(x)=\cos x  at x=0 is given as:

f'(0)= \lim_{h \to 0} \frac {f(0+h)-f(0)} h

⇒ f'(0)= \lim_{h \to 0} \frac {\cos(0+h)-\cos(0)} h

⇒ f'(0)= \lim_{h \to 0} \frac {\cos(h)-\cos(0)} h

⇒ f'(0)= \lim_{h \to 0} \frac {\cos(h)-1} h

∵ we can not find the limit of the above function f(x)=\cos x  by direct substitution as it gives 0/0 form (indeterminate form)

So we will simplify it to find the limit.

As we know that 1 – \cos x = 2 \sin2(x/2)

∴ f'(0)= \lim_{h \to 0} \frac {-(1-\cos(h))} h = - \lim_{h \to 0} \frac {2\sin^2(h/2)} h

Divide the numerator and denominator by 2 to get the form \sin x/x  for applying sandwich theorem and multiplying h in numerator and denominator to get the required form.

⇒ f'(0)= -\lim_{h \to 0}( \frac {\frac {2\sin^2(h/2)}2} {\frac {h^2}2}) \times h

⇒ f'(0)= -\lim_{h \to 0}( \frac {\sin(h/2)} {\frac {h}2}) \times \lim_{h \to 0}h

Using the formula: \lim_{x \to 0} \frac{\sin x}x=1

∴ f'(0)=-1 \times 0=0

Hence, derivative of f(x)=\cos x  at x=0 is 0

Question 6. Find the derivative of f(x) = \tan x  at x = 0

Solution:

Given: f(x)=\tan x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h  {where h is a small positive number}

Derivative of f(x)=\tan x  at x=0 is given as:

f'(0)= \lim_{h \to 0} \frac {f(0+h)-f(0)} h

⇒ f'(0)= \lim_{h \to 0} \frac {\tan(0+h)-\tan(0)} h

⇒ f'(0)= \lim_{h \to 0} \frac {\tan(h)-0} h

⇒ f'(0)= \lim_{h \to 0} \frac {\tan(h)} h

∴ Use the formula:  \lim_{x \to 0} \frac {\tan x} x=1  {sandwich theorem}

⇒ f'(0)=1

Hence, derivative of f(x)=\tan x  at x=0 is 1

Question 7(i). Find the derivatives of the following functions at the indicated points : \sin x  at x=\pi/2

Solution:

Given: f(x)= \sin x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h  {where h is a small positive number}

Derivative of f(x)=\sin x  at x=\pi/2  is given as:

f'(\pi/2)= \lim_{h \to 0} \frac {f(\pi/2+h)-f(\pi/2)} h

⇒ f'(\pi/2)= \lim_{h \to 0} \frac {\sin(\pi/2+h)-\sin(\pi/2)} h

⇒  f'(\pi/2)= \lim_{h \to 0} \frac {\sin(\pi/2+h)-1} h

⇒  f'(\pi/2)= \lim_{h \to 0} \frac {\cos(h)-1} h   {∵\sin(\pi/2+x)=\cos x

∵ we can not find the limit of the above function by direct substitution as it gives 0/0 form (indeterminate form)

So we will simplify it to find the limit.

As we know that

1 – \cos x = 2 \sin^2(x/2)

∴ f'(\pi/2)= \lim_{h \to 0} \frac {-(1-\cos(h))} h = - \lim_{h \to 0} \frac {2\sin^2(h/2)} h

Divide the numerator and denominator by 2 to get the form (sin x)/x for applying sandwich theorem and multiplying h in numerator and denominator to get the required form.

⇒ f'(\pi/2)= -\lim_{h \to 0}( \frac {\frac {2\sin^2(h/2)}2} {\frac {h^2}2}) \times h

⇒ f'(\pi/2)= -\lim_{h \to 0}( \frac {\sin(h/2)} {\frac {h}2}) \times \lim_{h \to 0}h

Using the formula:

\lim_{x \to 0} \frac{\sin x}x=1

∴ f'(\pi/2)=-1 \times 0=0

Hence, derivative of f(x)= \sin x  at x=\pi/2  is 0

Question 7(ii). Find the derivatives of the following functions at the indicated points : x at x=1

Solution:

Given: f(x)=x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h  {where h is a small positive number}

Derivative of f(x)=x at x=1 is given as:

f'(1)= \lim_{h \to 0} \frac {f(1+h)-f(1)} h

⇒ f'(1)= \lim_{h \to 0} \frac {(1+h)-1} h

⇒ f'(1)= \lim_{h \to 0} \frac {h} h

⇒ f'(1)= \lim_{h \to 0} 1 = 1

Hence, derivative of f(x)=x at x=1 is 1

Question 7(iii). Find the derivatives of the following functions at the indicated points : 2\cos x at x=\pi/2

Solution:

Given: f(x)= 2\cos x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h  {where h is a small positive number}

Derivative of f(x)= 2\cos x  at x=\pi/2  is given as:

f'(\pi/2)= \lim_{h \to 0} \frac {f(\pi/2+h)-f(\pi/2)} h

⇒ f'(\pi/2)= \lim_{h \to 0} \frac {2\cos(\pi/2+h)-2cos(\pi/2)} h

⇒ f'(\pi/2)= \lim_{h \to 0} \frac {-2\sin(h)} h {∵ \cos(\pi/2+x)=-\sin x }

∵ we can not find the limit of the above function by direct substitution as it gives 0/0 form (indeterminate form)

∴  f'(\pi/2)= -2\lim_{h \to 0} \frac {\sin h} h 

Using the formula:  \lim_{x \to 0} \frac{\sin x}x=1

∴ f'(\pi/2)=-2 \times 1=-2

Hence, derivative of f(x)= 2\cos x =-2

Question 7(iv). Find the derivatives of the following functions at the indicated points : \sin 2x  at x=\pi/2

Solution:

Given: f(x)= \sin  2x

By using the derivative formula,

f'(a)= \lim_{h \to 0} \frac {f(a+h)-f(a)} h  {where h is a small positive number}

Derivative of f(x)= \sin 2x  at x=\pi/2  is given as:

f'(\pi/2)= \lim_{h \to 0} \frac {f(\pi/2+h)-f(\pi/2)} h

⇒ f'(\pi/2)= \lim_{h \to 0} \frac {\sin 2(\pi/2+h)-sin 2(\pi/2)} h

⇒ f'(\pi/2)= \lim_{h \to 0} \frac {\sin(\pi+2h) - \sin \pi} h  {∵\sin(\pi+x)=-sinx }

⇒ f'(\pi/2)= \lim_{h \to 0} \frac {-\sin(2h) - 0} h

⇒ f'(\pi/2)= \lim_{h \to 0} \frac {-\sin(2h)} h

∵ we can not find the limit of the above function by direct substitution as it gives 0/0 form (indeterminate form)

Using the sandwich theorem  and multiplying 2 in numerator and denominator to apply the formula.

f'(\pi/2)= -\lim_{h \to 0} \frac {\sin(2h)} {2h} \times 2 = -2\lim_{h \to 0} \frac {\sin(2h)} {2h} 

Using the formula: \lim_{x \to 0} \frac{\sin x}x=1

∴ f'(\pi/2)=-2 \times 1=-2

Hence, derivative of f(x)=\sin 2x=-2