Mathematics Derivatives of implicit functions and Derivatives of inverse trigonometric functions For CBSE-NCERT
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### Topic covered

♦ Derivatives of implicit functions
♦ Derivatives of inverse trigonometric functions

### Derivatives of implicit functions :

=> Now we use chain rule to find derivatives of these functions.
Q 3115423369

Find (dy)/(dx)  if x – y = π.
Class 12 Chapter 5 Example 24
Solution:

One way is to solve for y and rewrite the above as

 y= x - pi

But then  (dy)/(dx) =1
Q 3115523460

Find (dy)/(dx), if y + sin y = cos x.
Class 12 Chapter 5 Example 25
Solution:

We differentiate the relationship directly with respect to x, i.e.,

 (dy)/(dx) + d/(dx) (sin y) = d/(dx) (cos x)

which implies using chain rule

 (dy)/(dx) + cos y * (dy)/(dx) = - sin x

This gives (dy)/(dx) = - (sin x)/(1+cos y)

where  y ≠ (2n + 1) π

### Derivatives of inverse trigonometric functions :

● We remark that inverse trigonometric functions are continuous functions, Now we use chain rule to find derivatives of these functions.

● E.g., : Find the derivative of f given by f (x) = tan^(–1) x assuming it exists.

Let y = tan^(–1) x.

Then, x = tan y.

=> Differentiating both sides w.r.t. x, we get

1=sec^2 y (dy)/(dx)

which implies that

(dy)/(dx) = 1/(sec^2 y) = 1/(tan^2 y) =1/(1+ (tan(tan^(-1) x))^2) =1/(1+x^2)
Q 3145623563

Find the derivative of f given by f (x) = sin^(–1) x assuming it exists.
Class 12 Chapter 5 Example 26
Solution:

Let y = sin^(–1) x. Then, x = sin y.
Differentiating both sides w.r.t. x, we get

1 = cos y (dy)/(dx)

which implies that  (dy)/(dx) = 1/(cos y) = 1/(cos (sin^(-1) x) )

Observe that this is defined only for cos y ≠ 0, i.e., sin^(–1) x ≠ - pi/2 , pi/2 , i.e., x ≠ – 1, 1,

i.e., x ∈ (– 1, 1).
To make this result a bit more attractive, we carry out the following manipulation.
Recall that for x ∈ (– 1, 1), sin (sin^(–1) x) = x and hence

cos^2 y = 1 – (sin y)^2 = 1 – (sin (sin^(–1) x))^2 = 1 – x^2

Also, since y ∈ ( -pi/2, pi/2) , cos y is positive and hence cos y = sqrt ( 1-x^2)

Thus, for x ∈ (– 1, 1),

 (dy)/(dx) = 1/(cos y ) = 1/(sqrt (1-x^2) )
Q 3105623568

Find the derivative of f given by f (x) = tan^(–1) x assuming it exists.
Class 12 Chapter 5 Example 27
Solution:

Let y = tan^(–1 )x. Then, x = tan y.
Differentiating both sides w.r.t. x, we get

1= sec^2 y (dy)/(dx)

which implies that

 (dy)/(dx) = 1/(sec^2 y) =1/(1+ tan^2 y) = 1/( 1+(tan( tan^(-1) x ))^2 ) = 1/(1+x^2)

Finding of the derivatives of other inverse trigonometric functions is left as exercise.
The following table gives the derivatives of the remaining inverse trigonometric functions
(Table 5.4):