♦ Derivatives of implicit functions

♦ Derivatives of inverse trigonometric functions

♦ Derivatives of inverse trigonometric 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

Class 12 Chapter 5 Example 24

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

Class 12 Chapter 5 Example 25

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 `

● 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)`

● 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

Class 12 Chapter 5 Example 26

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

Class 12 Chapter 5 Example 27

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):