Derivatives Of Functions In Parametric Forms

Given,

y%3D%5Csin%20%5E%7B2%7D%5Cleft(%5Ccot%20%5E%7B-1%7D%20%5Csqrt%7B%5Cfrac%7B1%2Bx%7D%7B1-x%7D%7D%5Cright)

Let

x%20%3D%5Ccos%202%20%5Ctheta%20%5Cdots

(i)

%5Ctherefore%20y%20%3D%5Csin%20%5E%7B2%7D%5Cleft%5B%5Ccot%20%5E%7B-1%7D%20%5Csqrt%7B%5Cfrac%7B1%2B%5Ccos%202%20%5Ctheta%7D%7B1-%5Ccos%202%20%5Ctheta%7D%7D%5Cright%5D

%5Cleft%5B%5Cbecause%201%2B%5Ccos%202%20%5Ctheta%3D2%20%5Ccos%20%5E%7B2%7D%20%5Ctheta%20%5Ctext%20%7B%20and%20%7D%201-%5Ccos%202%20%5Ctheta%3D2%20%5Csin%20%5E%7B2%7D%20%5Ctheta%5Cright%5D

%3D%20%5Csin%20%5E%7B2%7D%5Cleft%5B%5Ccot%20%5E%7B-1%7D%20%5Csqrt%7B%5Cfrac%7B2%20%5Ccos%20%5E%7B2%7D%20%5Ctheta%7D%7B2%20%5Csin%20%5E%7B2%7D%20%5Ctheta%7D%7D%5Cright%5D

%3D%20%5Csin%20%5E%7B2%7D%5Cleft%5B%5Ccot%20%5E%7B-1%7D(%5Ccot%20%5Ctheta)%5Cright%5D

%5CRightarrow%20y%3D%5Csin%20%5E%7B2%7D%20%5Ctheta

[Differentiating w.r.t.

%5Ctheta

, we get

%5D

%5CRightarrow%20%5Cfrac%7Bd%20y%7D%7Bd%20%5Ctheta%7D%3D2%20%5Csin%20%5Ctheta%20%5Ccos%20%5Ctheta

%5CRightarrow%20%5Cfrac%7Bd%20y%7D%7Bd%20%5Ctheta%7D%3D%5Csin%202%20%5Ctheta%20%5Cldots

(ii)
Differentiating E (i) w.r.t.

%5Ctheta

, we get

%5Cfrac%7Bd%20x%7D%7Bd%20%5Ctheta%7D%3D-2%20%5Csin%202%20%5Ctheta%20%5Cdots

(iii)

%5Cbecause%20%5Cfrac%7Bd%20y%7D%7Bd%20x%7D%3D%5Cfrac%7Bd%20y%20%2F%20d%20%5Ctheta%7D%7Bd%20x%20%2F%20d%20%5Ctheta%7D%3D%5Cfrac%7B%5Csin%202%20%5Ctheta%7D%7B-2%20%5Csin%202%20%5Ctheta%7D

[by Eqs. (ii) and (iii)]

%5CRightarrow%20%5Cfrac%7Bd%20y%7D%7Bd%20x%7D%3D-%5Cfrac%7B1%7D%7B2%7D

Step 1: Understanding the given function
We have:

y%20%3D%20%7C%5Csin%20x%20-%20%5Ccos%20x%7C.

To simplify this expression, we use the transformation:

%5Csin%20x%20-%20%5Ccos%20x%20%3D%20%5Csqrt%7B2%7D%20%5Csin%20%5Cleft(%20x%20-%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Cright).

Thus, the function can be rewritten as:

y%20%3D%20%5Cleft%7C%20%5Csqrt%7B2%7D%20%5Csin%20%5Cleft(%20x%20-%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Cright)%20%5Cright%7C.

Step 2: Finding points where $%5Csin(x%20-%20%5Cfrac%7B%5Cpi%7D%7B4%7D)%20%3D%200$

%5Csin(x%20-%20%5Cfrac%7B%5Cpi%7D%7B4%7D)%20%3D%200%20%5Cquad%20%5CRightarrow%20%5Cquad%20x%20-%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%3D%200.

Solving for

x

x%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D.

Step 3: Evaluating the area integral
The function changes sign at

x%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D

. Therefore, we split the integral:

A%20%3D%20%5Cint_0%5E%7B%5Cpi%2F4%7D%20(%5Ccos%20x%20-%20%5Csin%20x)%20%5C%2C%20dx%20%2B%20%5Cint_%7B%5Cpi%2F4%7D%5E%7B%5Cpi%2F2%7D%20(%5Csin%20x%20-%20%5Ccos%20x)%20%5C%2C%20dx.

Using standard integration:

%5Cint%20(%5Ccos%20x%20-%20%5Csin%20x)%20dx%20%3D%20%5Csin%20x%20%2B%20%5Ccos%20x.

Evaluating in

%5B0%2C%20%5Cpi%2F4%5D

%5Cleft%5B%20%5Csin%20x%20%2B%20%5Ccos%20x%20%5Cright%5D_0%5E%7B%5Cpi%2F4%7D%20%3D%20(%5Csin%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2B%20%5Ccos%20%5Cfrac%7B%5Cpi%7D%7B4%7D)%20-%20(0%20%2B%201).

%3D%20(%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%2B%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)%20-%201%20%3D%20%5Csqrt%7B2%7D%20-%201.

Similarly, for

%5B%5Cpi%2F4%2C%20%5Cpi%2F2%5D

%5Cleft%5B%20%5Csin%20x%20%2B%20%5Ccos%20x%20%5Cright%5D_%7B%5Cpi%2F4%7D%5E%7B%5Cpi%2F2%7D%20%3D%20(1%20%2B%200)%20-%20(%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%2B%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D).

%3D%201%20-%20%5Csqrt%7B2%7D%20%2B%201.

Adding both areas:

A%20%3D%202%20(%5Csqrt%7B2%7D%20-%201).

Step 4: Conclusion
Thus, the correct answer is:

2(%5Csqrt%7B2%7D%20-%201).

Chapter Questions
5 MCQs