9. Differentiate (Exercise) 9.2

Q1. Differentiate the following w.r.t.x :

1. y = `x^(4/3) + "e"^x - sinx`

y = `x^(4/3) + "e"^x - sinx`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x)(x^(4/3) + "e"^x - sinx)`

∴ `("d"y)/("d"x) = "d"/("d"x) (x^(4/3)) + "d"/("d"x)("e"^x) - "d"/("d"x)(sinx)`

= `4/3x^(4/3 - 1) + "e"^x - cos x`

= `4/3 x^(1/3) + "e"^x - cos x`

 2. y = `sqrt(x) + tan x - x^3`

Let y = `sqrt(x) + tan x - x^3`

∴ `("d"y)/("d"x) = "d"/("d"x) (sqrt(x) + tan x - x^3)`

= `"d"/("d"x)(sqrt(x)) + "d"/("d"x) (tanx) - "d"/("d"x)(x^3)`

= `1/(2sqrt(x)) + sec^2x - 3x^2`.

3. y = `log x - "cosec"  x + 5^x - 3/(x^(3/2))`

y = `log x - "cosec"  x + 5^x - 3/(x^(3/2))`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x)(log x - "cosec"  x + 5^x - 3/(x^(3/2)))`

∴ `("d"y)/("d"x) = "d"/("d"x)(logx) - "d"/("d"x) ("cosec"  x) + "d"/("d"x)(5^x) - 3"d"/("d"x)(x^(-3/2))`

= `1/x - (- "cosec"  x cot  x) + 5^x log 5 - 3(-3/2 x^((-3)/2 - 1))`

= `1/x - (- "cosec"  x cot  x) + 5^x log 5 - 3(-3/2 x^((-5)/2))`

= `1/x + "cosec"  x cot  x + 5^x log 5 + 9/(2x^(5/2))`

4. y = `x^(7/3) + 5x^(4/5) - 5/(x^(2/5))`

Let y = `x^(7/3) + 5x^(4/5) - 5/(x^(2/5))`

∴ `("d"y)/("d"x) = "d"/("d"x) [x^(7/3) + 5x^(4/5) - 5/(x^(2/5))]`

= `"d"/("d"x) (x^(7/3)) + "d"/("d"x)(5x^(4/5)) - "d"/("d"x)(5x^(-2/5))`

= `7/3x^(4/3) + 5 xx 4/5x^(-1/5)  -5 xx (-2/5)x^(-7/3)`

= `7/3x^(4/3) + 4/(x^(1/5)) + 2/(x^(7/5))`

5. y = `7^x + x^7 - 2/3 xsqrt(x) - logx + 7^7`

y = `7^x + x^7 - 2/3 xsqrt(x) - logx + 7^7`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x) (7^x + x^7 - 2/3 xsqrt(x) - log x + 7^7)`

∴ `("d"y)/("d"x) = "d"/("d"x) (7^x) + "d"/("d"x) (x^7) - 2/3 "d"/("d"x) (x^(3/2)) - "d"/("d"x) (logx) + "d"/("d"x) 7^7`

= `7^x log 7 + 7x^6 - 2/3 xx 3/2 x^(3/2 - 1) - 1/x + 0`

= `7^x log 7 + 7x^6 - x^(1/2) - 1/x`

= `7^x log 7 + 7x^6 - sqrt(x) - 1/x`

6. y = `3 cotx - 5"e"^x + 3logx - 4/(x^(3/4))`

Let y = `3 cotx - 5"e"^x + 3logx - 4/(x^(3/4))`

∴ `("d"y)/("d"x) = "d"/("d"x) [3 cot x - 5"e"^x + 3 log x - 4x^(-3/4)]`

= `"d"/("d"x) (3 cot x) - "d"/("d"x) (5"e"^x) + "d"/("d"x) (3 log x) - "d"/("d"x) (4x^(-3/4))`

= `3 "d"/("d"x) (cot x) - 5 "d"/("d"x) ("e"^x) + 3"d"/("d"x) (log x) - 4 "d"/("d"x) (x^(-3/4))`

= `3 xx (- "cosec"^2x) - 5"e"^x + 3 xx 1/x - 4 xx (-3/4)x^(-7/4)`

= `- 3  "cosec"^2x - 5"e"^x + 3/x + 3/(x^(7/4)`

Q2.Differentiate the following w.r.t.x. :

1. y = x⁵ tan x

y = x⁵ tan x

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x)(x^5 tanx)`

= `"x"^5 "d"/("d"x) tanx + tanx "d"/("d"x) x^5`

= x⁵ sec²x + tan x (5x⁴)

= x⁴ (x sec²x + 5tan x)

2. y = x³ log x

Let y = x³ log x

∴ `("d"y)/("d"x) = "d"/("d"x) [x^3 log x]`

= `x^3 "d"/("d"x) (logx) + (logx) "d"/("d"x) (x^3)`

= `x^3 (1/x) + (log x)(3x^2)`

= x² + 3x² log x

= x² (1 + 3 log x)

3. y = (x² + 2)² sin x

y = (x² + 2)² sin x

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x) [(x^2 + 2)^2 sin x]`

`("d"y)/("d"x) = (x^2 + 2)^2 "d"/("d"x) sinx + sinx ["d"/("d"x) (x^2 + 2)^2]`

= `(x^2 + 2)^2 cos x + sin x "d"/("d"x) (x^4 + 4x^2 + 4)`

= `(x^2 + 2)^2 cos x + sin x(4x^3 + 8x)`

= `(x^2 + 2)^2 cos x + sinx* (4x)(x^2 + 2)`

= `(x^2 + 2)[(x^2 + 2) cosx + 4x sinx]`

4. y = e^x logx

y = e^x logx

∴ `("d"y)/("d"x) = "d"/("d"x)["e"^x log x]`

= `"e"^x "d"/("d"x) (log x) + (log x)"d"/("d"x) ("e"^x)`

= `"e"^x (1/x) + (log x)("e"^x)`

= `"e"^x (1/x + log x)`

5. y = `x^(3/2)  "e"^xlogx`

y = `x^(3/2) ( "e"^xlogx)`

Differentiate with respect to x, we get

`("d"y)/("d"x) = "d"/("d"x) (x^(3/2) ("e"^x log x))`

`("d"y)/("d"x) = x^(3/2) "d"/("d"x) ("e"^x log x) + ("e"^x log x) "d"/("d"x)(x^(3/2))`

= `x^(3/2) ["e"^x "d"/("d"x) log x + log x "d"/("d"x) "e"^x] + ("e"^x log x)(3/2 x^(3/2 - 1))`

= `x^(3/2) ["e"^x (1/x) + log x ("e"^x)] + ("e"^x log x) (3/2 x^(1/2))`

= `"e"^x  x^(3/2) (1/x + log x) + ("e"^x log x)(3/2 x^(1/2))`

= `x^(1/2) "e"^x[x(1/x + log x) + (log x)(3/2)]`

= `sqrt(x)"e"^x [1 + x log x + 3/2 log x]`

6. y = log e^_x3 log x³ 

Let y = (log e^_x3) (log x³) = (x³ log e)(log x³)

= x³ log x³    ...[∵ log e = 1]

= x³ (3 log x)

= 3x³ log x

∴ `("d"y)/("d"x) = "d"/("d"x) [3x^3 log x]`

= `3"d"/("d"x) (x^3 log x)`

= `3[x^3 "d"/("d"x) (log x) + (log x) "d"/("d"x) (x^3)]`

= `3[x^3 (1/x) + (log x)(3x^2)]`

= 3x² + 9x² log x

= 3x² (1 + 3 log x)

Q3. Differentiate the following w.r.t.x. :

1. y = `x^2sqrt(x) + x^4logx`

y = `x^2sqrt(x) + x^4logx`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x) (x^2 sqrt(x) + x^4 log x)`

`("d"y)/("d"x) = "d"/("d"x) (x^(5/2)) + "d"/("d"x) (x^4 log * x)`

= `5/2x^(5/2 - 1) + [x^4 "d"/("d"x) log x + log x("d"/("d"x) x^4)]`

= `5/2x^(3/2) + [x^4 (1/x) + log x(4x^3)]`

= `5/2x^(3/2) + x^3 (1 + 4logx)`

2. y = `"e"^xsecx - x^(5/3) log x`

Let y = `"e"^xsecx - x^(5/3) log x`

∴ `("d"y)/("d"x) = "d"/("d"x)["e"^x sec x - x^(5/3) log x]`

= `"d"/("d"x)["e"^x sec x] - "d"/("d"x)[x^(5/3) log x]`

= `"e"^x "d"/("d"x) (sec x) + sec x "d"/("d"x) ("e"^x) - [x^(5/3) "d"/("d"x) (log x) + (log x) "d"/("d"x) (x^(5/3))]`

= `"e"^x (sec x tan x) + sec x("e"^x) - x^(5/3) (1/x) - (log x) (5/3 x^(2/3))`

= `"e"^x sec x (tan x + 1) - x^(2/3) (1 + 5/3 log x)`

3. y = `x^4 + x sqrt(x) cos x - x^2"e"^x`

y = `x^4 + x sqrt(x) cos x - x^2"e"^x`

y = `x^4 + x^(3/2) cos x - x^2"e"^x`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x) (x^4 + x^(3/2) cos x - x^2 "e"^x)`

`("d"y)/("d"x) = "d"/("d"x) (x^4) + "d"/("d"x) (x^(3/2) cos x) - "d"/("d"x) (x^2"e"^x)`

= `4x^3 + x^(3/2) "d"/("d"x) cos x + cos x "d"/("d"x) (x^(3/2)) - [x^2 ("d"/("d"x) "e"^x) + "e"^x ("d"/("d"x) x^2)]`

= `4x^3 + x^(3/2) (- sin x) + cos x (3/2x^(1/2)) - [x^2 "e"^x + "e"^x (2x)]`

= `4x^3 - x^(3/2) sin x + 3/2 sqrt(x) cos x - x"e"^x (x + 2)`

4. y = (x³ – 2) tan x – x cos x + 7^x. x⁷

Let y = (x³ – 2) tan x – x cos x + 7^x. x⁷  

∴ `("d"y)/("d"x) = "d"/("d"x) [(x^3 - 2) tan x - x cos x + 7^x*x^7]`

= `"d"/("d"x)[(x^3 - 2) tan x] - "d"/("d"x)[x cos x] + "d"/("d"x) [7^x*x^7]`

= `(x^3 - 2) "d"/("d"x) (tan x) + (tan x) "d"/("d"x) (x^3 - 2) - [x "d"/("d"x) (cos x) + (cos x) "d"/("d"x) (x)] + 7^x "d"/("d"x) (x^7) + x^7 "d"/("d"x)(7^x)`

= (x³ – 2)(sec²x) + (tan x)(3x² – 0) – [x(– sin x) + (cos x)(1)] + 7^x (7x⁶) + x⁷ (7^x log 7)

= (x³ – 2)sec²x + 3x² tan x + x sin x – cos x + 7^x. x⁶ [7 + x log 7]

5. y = `sinx logx + "e"^x cos x - "e"^x sqrt(x)`

Let y = `sinx logx + "e"^x cos x - "e"^x sqrt(x)`

∴ `("d"y)/("d"x) = "d"/("d"x) [sinx * log x + "e"^x cos x - "e"^x sqrt(x)]`

= `"d"/("d"x) [sinx * logx] + "d"/("d"x) ["e"^x cos x] - "d"/("d"x) ["e"^x sqrt(x)]`

= `(sin x) "d"/("d"x) (log x) + (log x) "d"/("d"x) (sin x) + "e"^x "d"/("d"x) (cos x) + cos x "d"/("d"x) ("e"^x) - ["e"^x "d"/("d"x) (sqrt(x)) + sqrt(x) "d"/("d"x) ("e"^x)]`

= `(sin x) (1/x) + (log x)(cos x) + "e"^x (- sin x) + cos x ("e"^x) - ["e"^x (1/(2sqrt(x))) + sqrt(x)("e"^x)]`

= `sinx/x + (cos x)(log x) - "e"^x sin x + "e"^x cos x - "e"^x/(2sqrt(x)) - "e"^xsqrt(x)`

= `sinx/x + (cos x)(log x) + "e"^x(- sinx + cos x - 1/(2sqrt(x)) - sqrt(x))`

= `sinx/x + (cos x)(log x) + "e"^x [- sin x + cos x - ((1 + 2x)/(2sqrt(x)))]`

6. y = `"e"^x tanx + cos x log x - sqrt(x)  5^x`

y = `"e"^x tanx + (cos x) (log x) - (sqrt(x)) 5^x`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x) ("e"^x tan x) + "d"/("d"x) (cos x log x) - "d"/("d"x) [(sqrt(x)) 5^x]`

= `("e"^x "d"/("d"x) tan x + tan x "d"/("d"x) "e"^x) + [cos x "d"/("d"x) (log x) + log x "d"/("d"x) (cos x)] - [sqrt(x)"d"/("d"x )(5^x) + 5^x "d"/("d"x) sqrt(x)]`

= `"e"^x sec^2x + tan x ("e"^x) + cos x (1/x) + logx(- sin x) - [sqrt(x)(5^x log 5) + 5^x 1/(2sqrt(x))]`

= `"e"^x (sec^2x + tan x) + cos x/x - sinx logx - [(2x  5^x log 5 + 5^x)/(2sqrt(x))]`

= `"e"^x (sec^2x + tan x) + cosx/x - sinx logx - 5^x((2x log5 + 1)/(2sqrt(x)))`

Q4. Differentiate the following w.r.t.x. :

1. y = `(x^2 + 3)/(x^2 - 5)`

Let y = `(x^2 + 3)/(x^2 - 5)`

∴ `("d"y)/("d"x) = "d"/("d"x) [(x^2 + 3)/(x^2 - 5)]`

= `((x^2 - 5) "d"/("d"x) (x^2 + 3) - (x^2 + 3) "d"/("d"x) (x^2 - 5))/(x^2 - 5)^2`

= `((x^2 - 5)(2x + 0) - (x^2 + 3)(2x - 0))/(x^2 - 5)^2`

= `(2x[(x^2 - 5) - (x^2 + 3)])/(x^2 - 5)^2`

= `(- 16x)/(x^2 - 5)^2`

2. y = `(sqrt(x) + 5)/(sqrt(x) - 5)`

y = `(sqrt(x) + 5)/(sqrt(x) - 5)`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x) ((sqrt(x) + 5)/(sqrt(x) - 5))`

= `((sqrt(x) - 5) "d"/("d"x) (sqrt(x) + 5) - (sqrt(x) + 5) "d"/("d"x) (sqrt(x) - 5))/(sqrt(x) - 5)^2`

= `((sqrt(x) - 5) (1/(2sqrt(x))) - (sqrt(x) + 5)(1/(2sqrt(x))))/(sqrt(x) - 5)^2`

= `(1/(2sqrt(x)) (sqrt(x) - 5 - sqrt(x) - 5))/(sqrt(x) - 5)^2`

= `(1/(2sqrt(x)) (-10))/(sqrt(x) - 5)^2`

= `(-5)/(sqrt(x) (sqrt(x) - 5)^2`

3. y = `(x"e"^x)/(x + "e"^x)`

Let y = `(x"e"^x)/(x + "e"^x)`

∴ `("d"y)/("d"x) = "d"/("d"x) [(x"e"^x)/(x + "e"^x)]`

= `((x + "e"^x) "d"/("d"x) (x"e"^x) - x"e"^x "d"/("d"x) (x + "e"^x))/(x + "e"^x)^2`

= `((x + "e"^x) [x "d"/("d"x) ("e"^x) + "e"^x "d"/("d"x) (x)] - x"e"^x (1 + "e"^x))/(x + "e"^x)^2`

= `((x + "e"^x) [x("e"^x) + "e"^x (1)] - x"e"^x - x"e"^(2x))/(x + "e"^x)^2`

= `(x^2"e"^x + x"e"^x + x"e"^(2x) + "e"^(2x) - x"e"^x - x"e"^(2x))/(x + "e"^x)^2`

= `(x^2"e"^x + "e"^(2x))/(x + "e"^x)^2`

= `("e"^x (x^2 + "e"^x))/(x + "e"^x)^2`

4. y = `(x"e"^x)/(x + "e"^x)`

Let y = `(x"e"^x)/(x + "e"^x)`

∴ `("d"y)/("d"x) = "d"/("d"x) [(x"e"^x)/(x + "e"^x)]`

= `((x + "e"^x) "d"/("d"x) (x"e"^x) - x"e"^x "d"/("d"x) (x + "e"^x))/(x + "e"^x)^2`

= `((x + "e"^x) [x "d"/("d"x) ("e"^x) + "e"^x "d"/("d"x) (x)] - x"e"^x (1 + "e"^x))/(x + "e"^x)^2`

= `((x + "e"^x) [x("e"^x) + "e"^x (1)] - x"e"^x - x"e"^(2x))/(x + "e"^x)^2`

= `(x^2"e"^x + x"e"^x + x"e"^(2x) + "e"^(2x) - x"e"^x - x"e"^(2x))/(x + "e"^x)^2`

= `(x^2"e"^x + "e"^(2x))/(x + "e"^x)^2`

= `("e"^x (x^2 + "e"^x))/(x + "e"^x)^2`

5. y = `(x^2 sin x)/(x + cos x)`

Let y = `(x^2 sin x)/(x + cos x)`

∴ `("d"y)/("d"x) = "d"/("d"x) [(x^2 sin x)/(x + cos x)]`

= `((x + cos x) "d"/("d"x) (x^2 sin x) - (x^2 sin x) "d"/("dx") (x + cos x))/(x + cos x)^2`

= `((x + cos x) [x^2 "d"/("d"x) (sin x) + (sin x) "d"/("d"x) (x^2)] - (x^2 sin x) (1 - sin x))/(x + cos x)^2`

= `((x + cos x) [x^2 (cos x) + (sin x)(2x)] - x^2 sin x + x^2 sin^2x)/(x + cos x)^2`

= `(x^3 cos x + 2x^2 sin x + x^2 cos^2x + 2x sin x cos x + x^2 sin^2x - x^2 sinx)/(x + cos x)^2`

= `(x^3 cos x + x^2 sin x + x^2 cos^2x + x^2 sin^2x + x sin 2x)/(x + cos x)^2`

= `(x^2[x cos x + sin x + (cos^2x + sin^2x)] + x sin 2x)/(x + cos x)^2`

= `(x^2[x cos x + sin x + 1] + x sin 2x)/(x + cos x)^2`

6. y = `(5"e"^x - 4)/(3"e"^x - 2)`

y = `(5"e"^x - 4)/(3"e"^x - 2)`

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x) ((5"e"^x - 4)/(3"e"^x - 2))`

`("d"y)/("d"x) = ((3"e"^x - 2) "d"/("d"x) (5"e"^x - 4) - (5"e"^x - 4) "d"/("d"x) (3"e"^x - 2))/(3"e"^x - 2)^2`

= `((3"e"^x - 2)(5 "d"/("d"x) "e"^x - "d"/("d"x) 4) - (5"e"^x - 4)(3 "d"/("d"x) "e"^x - "d"/("d"x) 2))/(3"e"^x - 2)^2`

= `((3"e"^x - 2)(5"e"^x - 0) - (5"e"^x - 4)(3"e"^x - 0))/(3"e"^x - 2)^2`

= `(15("e"^x)^2 - 10"e"^x - 15("e"^x)^2 + 12"e"^x)/(3"e"^x - 2)^2`

= `(2"e"^x)/(3"e"^x - 2)^2`

Q5. If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12 then find f(x)

Let f(x) = ax² + bx + c;  a ≠ 0, a, b, c ∈ R

f'(x) = `"d"/("d"x) ("a"x^2 + "b"x + "c")` = a × 2x + b × 1 + 0 = 2ax + b

It is given that f(0) = 3, f'(2) = 2 and f'(3) = 12

But f(0) = a(0) + b(0) + c = c, f'(2) = 2a(2) + b = 4a +b and f'(3) = 2a(3) + b = 6a + b

∴ c = 3

4a + b = 2     ...(1)

and 6a + b = 12    ...(2)

Subtracting (1) from (2), we get 2a = 10

∴ a = 5

∴ from (1), 4(5) + b = 2

∴ b = 2 – 20 = – 18

∴ f(x) = 5x² – 18x + 3

2. If f(x) = a sin x – b cos x, `"f'"(pi/4) = sqrt(2) and "f'"(pi/6)` = 2, then find f(x)

f(x) = a sin x – b cos x

Differentiating w.r.t. x, we get

f'(x) = a cos x – b (– sin x)

f'(x) = a cos x + b sin x

∴ `"f'" (pi/4) = "a" cos (pi/4) + "b" sin (pi/4)`

But, `"f'"(pi/4) = sqrt(2)`   ...(given)

∴ `"a" cos  pi/4 + "b" sin  pi/4 = sqrt(2)`

∴ `"a"(1/sqrt(2)) + "b" (1/sqrt(2)) = sqrt(2)`

∴ `("a" + "b")/(sqrt(2)) = sqrt(2)`

∴ a + b = 2   ...(i)

Also, `"f'"(pi/6) = "a" cos  pi/6 + "b" sin  pi/6`

But, `"f'" (pi/6)` = 2   ...(given)

∴ `"a" cos  pi/6 + "b" sin  pi/6` = 2

∴ `"a"(sqrt(3))/2 + "b"(1/2)` = 2

∴ `"a" sqrt(3) + "b"` = 4   ...(ii)

equation (ii) – equation (i), we get

`"a" sqrt(3) - "a"` = 2

∴ `"a"(sqrt(3) - 1)` = 2

∴ a = `2/(sqrt(3) - 1) = (2(sqrt(3) + 1))/(3 - 1)`

∴ a = `sqrt(3) + 1`

Substituting a = `sqrt(3) + 1` in equation (i), we get

`sqrt(3) + 1 + "b"` = 2

`"b" = 1 - sqrt(3)`

Now, f(x) = a sin x – b cos x

∴ f(x) = `(sqrt(3) + 1) sin x + (sqrt(3) - 1) cos x`

Q6. Fill in the blanks:

1.y = e^x .tan x

Differentiating w.r.t.x

`("d"y)/("d"x) = "d"/("d"x)("e"^x tan x)`

= `square "d"/("d"x) tanx + tan x "d"/("d"x) square`

= `square  square + tan x  square`

= `"e"^x [square  + square]`

y = e^x .tan x

Differentiating w.r.t.x

`("d"y)/("d"x) = "d"/("d"x)("e"^x tan x)`

= `"e"^x "d"/("d"x) tanx + tan x "d"/("d"x) "e"^x`

= `"e"^x sec^2x + tan x  "e"^x`

= `"e"^x [sec^2x + tan x]`

2. y = `sinx/(x^2 + 2)`

Differentiating. w.r.t.x.

`("d"y)/("d"x) = (square "d"/("d"x) (sin x) - sin x "d"/("dx) square)/(x^2 + 2)^2`

= `(square  square - sin x  square)/(x^2 + 2)^2`

= `(square - square)/(x^2 + 2)^2`

y = `sinx/(x^2 + 2)`

Differentiating. w.r.t.x., we get

`("d"y)/("d"x) = (x^2+ 2 "d"/("d"x) (sin x) - sin x "d"/("dx) x^2 + 2)/(x^2 + 2)^2`

= `((x^2 + 2)  (cos x) - sin x  (2x))/(x^2 + 2)^2`

= `((x^2 cos x + 2 cos x) - 2x sin x)/(x^2 + 2)^2`

3. y = (3x² + 5) cos x

Differentiating w.r.t.x

`("d"y)/("d"x) = "d"/("d"x) [(3x^2 + 5) cos x]`

= `(3x^2 + 5) "d"/("d"x) [square] + cos x  "d"/("d"x) [square]`

= `(3x^2 + 5) [square] + cos x  [square]`

∴ `(dx)/("d"y) = (3x^2 + 5) [square] + [square] cos x`

y = (3x² + 5) cos x

Differentiating w.r.t.x,

`("d"y)/("d"x) = "d"/("d"x) [(3x^2 + 5) cos x]`

= `(3x^2 + 5) "d"/("d"x) [cos x] + cos x  "d"/("d"x) [3x^2 + 5]`

= `(3x^2 + 5) [- sin x] + cos x  [6x + 0]`

∴ `(dx)/("d"y) = (3x^2 + 5) [- sin x] + [6x] cos x`

4. Differentiate tan x and sec x w.r.t.x. using the formulae for differentiation of `"u"/"v" and 1/"v"` respectively

(i) Let y = `tan x = sin x/cos x`

∴ `("d"y)/("d"x) = "d"/("d"x) ((sin x)/(cos x))`

= `((cos x) "d"/("d"x) (sin x) - (sin x) "d"/("d"x) (cos x))/(cos^2x)`

= `((cos x)(cos x) - (sin x)(- sin x))/(cos^2x)`

= `(cos^2x + sin^2x)/(cos^2x)`

= `1/(cos^2x)`

= sec²x

(ii) Let y = sec x = `1/cos x`

∴ `("d"y)/("d"x) = "d"/("d"x) [1/cos x]`

= `((cos x) "d"/("d"x) (1) - (1) "d"/("d"x) (cos x))/(cos^2x)`

= `((cos x)(0) - (- sin x))/(cos^2x)`

= `sinx/cos^2x`

= `(1/(cos x))((sin x)/(cos x))`

= sec x tan x