differentiate w. r. t x
Answers
Step-by-step explanation:
Method 1:
Let y=(x
3
−2x−1)
5
Put u=x
3
−2x−1. Then y=u
5
∴
du
dy
=
du
d
(u
5
)=5u
4
=5(x3−2x−1)4 and
dx
du
=
dx
d
(x
3
−2x−1)
=3×2−2×1−0
=3×2−2
∴
dx
dy
=
du
dy
×
dx
du
=5(x
3
−2x−1)
4
(3x
2
−2)
=5(3x
2
−2)(x
3
−2x−1)
4
Method 2:
Let y=(x
3
−2x−1)
5
Differentiating w.r.t. x, we get
dx
dy
=
dx
d
(x
3
−2x−1)
5
=5(x
3
−2x−1)4×
dx
d
(x
3
−2x−1)
=5(x
3
−2x−1)
4
×(3x
2
−2×1−0)
=5(3x
2
−2)(x
3
−2x−1)
4
Note: Out of the two methods given above, we will use Method 2 for solving the remaining problems.
Answer:
1/ (2 tan x/2 sec^2 x/2) + cosec^2 x (log(1 + sin x) - cos x cot x / (1 + sin x) - 1
Step-by-step explanation:
y = log tan x/2 - cot x log(1 + sinx) - x
We use the Chain and Product Rules:-
dy/dx = 1 / tan x/2 * 1/2 sec^2 x/2 - ( -cosec^2 x * (log(1 + sin x) + cot x * 1/1 + sinx) * cosx) - 1
= 1/ (2 tan x/2 sec^2x/2) + cosec^2x (log(1 + sin x) - cos x cot x / (1 + sinx) - 1