Решения NCERT для класса 11 - Глава 13 Пределы и производные - Разное упражнение по главе 13 | Набор 2

Найдите производную следующих функций (следует понимать, что a, b, c, d, p, q, r и s — фиксированные ненулевые константы, а m и n — целые числа):

Вопрос 16:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

Вопрос 17:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

Вопрос 18:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

Вопрос 19: sin ⁿ x

Решение:

f(x) = sinⁿ x

When n = 1,

f(x) = sin x

When n = 2,

f(x) = sin² x = sin x sin x 

Using the product rule, we have

(uv)’ = uv’+vu’

f"(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

f(x) = sin³ x = sin² x sin x

Using the product rule, we have

(uv)’ = uv’+vu’

Pattern w.r.t n is seen here, as follows

Let’s check this statement.

For P(n) = n sin^n-1x cos x

For P(1),

P(1) = 1 sin^1-1x cos x = cos x. Which is true.

n=k

n = k+1

Using the product rule, we have

(uv)’ = uv’+vu’

= (sin^k x)  + (sin x) 

= (sin^k x) (cos x) + (sin x) (k sin^k-1 x cos x)

= (sin^k x) (cos x)[k+1]

Hence proved for P(k+1).

So, is true.

Вопрос 20:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Вопрос 21:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

Let’s take g(x) = sin (x+a)

g(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

Hence, 

Using the trigonometric identity,

cos A cos B + sin A sin B = cos (A-B)

Вопрос 22: x ⁴ (5sin x – 3cos x)

Решение:

f(x) = x⁴ (5sin x – 3cos x)

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

f"(x) = (x⁴) [(5 cos x) – (3 (- sin x))] + (5sin x – 3cos x)(4x³)

f"(x) = (x⁴) [(5 cos x) + (3 sin x)] + (5sin x – 3cos x)(4x³)

f"(x) = (x³) [5x cos x + 3x sin x + 20sin x – 12 cos x]

Вопрос 23: (x ² +1) cos x

Решение:

f(x) = (x²+1) cos x

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

f"(x) = (x²+1) (- sin x) + (cos x)[(2x^2-1)+0]

f"(x) = -x² sin x- sin x + 2x cos x

Вопрос 24:

Решение:

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Вопрос 25: (x + cos x)(x – тангенс x)

Решение:

f(x) = (x + cos x)(x – tan x)

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

g"(x) = sec²x

Hence, 

f"(x) = (x + cos x)  + (x – tan x)[1 + (- sin x)]

f"(x) = (x + cos x) [1 – (sec² x)] + (x – tan x)[1 – sin x]

f"(x) = (x + cos x) [tan² x] + (x – tan x)[1 – sin x]

f"(x) = tan² x(x + cos x) + (x – tan x)[1 – sin x]

Вопрос 26:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Вопрос 27:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Вопрос 28:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

g"(x) = sec²x

Hence, 

Вопрос 29: (x + sec x) (x-tan x)

Решение:

f(x) = (x + sec x) (x-tan x)

Taking derivative both sides,

Using the product rule, we have

(uv)’ = uv’ + vu’

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Let’s take g(x) = tan x

From the first principle,

Using the trigonometric identity,

sin a cos b – cos a sin b = sin (a-b)

g"(x) = sec²x

Now, let’s take h(x) = sec x = 

h(x+h) =

From the first principle,

Using the trigonometric identity,

cos a – cos b = -2 sin   sin 

Multiply and divide by 2, we have

Hence, 

Вопрос 30:

Решение:

Taking derivative both sides,

Using the quotient rule, we have

As, the derivative of xⁿ is nx^n-1 and derivative of constant is 0.

Let’s take, g(x) = sinn x

When n = 1,

g(x) = sin x

When n = 2,

Using the product rule, we have

(uv)’ = uv’+vu’

g"(x) = (sin x) (cos x) + (sin x) (cos x) = 2 sin x cos x

When n = 3,

g(x) = sin³ x = sin² x sin x

Using the product rule, we have

(uv)’ = uv’+vu’

Pattern w.r.t n is seen here, as follows

Let’s check this statement.

For 

For P(1),

. Which is true.

n=k

n = k+1

Using the product rule, we have

(uv)’ = uv’+vu’

Hence proved for P(k+1).

So,  is true.

So, the given equation will be