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# Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Miscellaneous Exercise on Chapter 13 | Set 1

• Last Updated : 30 Apr, 2021

### (i) -x

Solution:

f(x) = -x

f(x+h) = -(x+h)

From the first principle, f'(x) = -1

### (ii) (-x)-1

Solution:

f(x) = (-x)-1 f(x+h) = (-(x+h))-1 From the first principle, ### (iii) sin(x+1)

Solution:

f(x) = sin(x+1)

f(x+h) = sin((x+h)+1)

From the first principle, Using the trigonometric identity,

sin A – sin B = 2 cos sin  Multiply and divide by 2, we have f'(x) = cos (x+1) (1)

f'(x) = cos (x+1)

### (iv) Solution:

Here, From the first principle, Using the trigonometric identity,

cos a – cos b = -2 sin sin  Multiplying and diving by 2, ### Question 2: (x+a)

Solution:

f(x) = x+a

Taking derivative both sides, As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 3: Solution: Taking derivative both sides, Using the product rule, we have

(uv)’ = uv’+u’v As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 4: (ax+b) (cx+d)2

Solution:

f(x) = (ax+b) (cx+d)2

Taking derivative both sides, Using the product rule, we have

(uv)’ = uv’+u’v As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 5: Solution: Taking derivative both sides, Using the quotient rule, we have As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 6: Solution: Taking derivative both sides, Using the quotient rule, we have As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 7: Solution: Taking derivative both sides, Using the quotient rule, we have As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 8: Solution: Taking derivative both sides, Using the quotient rule, we have As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 9: Solution: Taking derivative both sides, Using the quotient rule, we have As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 10: Solution: Taking derivative both sides, As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 11: Solution: Taking derivative both sides, As, the derivative of xn is nxn-1 and derivative of constant is 0. ### Question 12: (ax+b)n

Solution:

f(x) = (ax+b)n

f(x+h) = (a(x+h)+b)n

f(x+h) = (ax+ah+b)n

From the first principle, Using the binomial expansion, we have ### Question 13: (ax+b)n (cx+d)m

Solution:

f(x) = (ax+b)n (cx+d)m

Taking derivative both sides, Using the product rule, we have

(uv)’ = uv’+u’v Let’s take, g(x) = (cx+d)m g(x+h) = (c(x+h)+d)m

g(x+h) = (cx+ch+d)m

From the first principle, Using the binomial expansion, we have So, as ### Question 14: sin (x + a)

Solution:

f(x) = sin(x+a)

f(x+h) = sin((x+h)+a)

From the first principle, Using the trigonometric identity,

sin A – sin B = 2 cos sin  Multiply and divide by 2, we have ### Question 15: cosec x cot x

Solution:

f(x) = cosec x cot x

Taking derivative both sides, Using the product rule, we have

(uv)’ = uv’+u’v f'(x) = cot x (-cot x cosec x) + (cosec x) (-cosec2 x)

f'(x) = – cot2 x cosec x – cosec3 x

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