Question

find the derivative of cosinverse (4xcube-3x)​

Answer 1

Answer:

- 3 / √(1 - x² )

Step-by-step explanation:

To find---> Derivative of Cos⁻¹ ( 4x³ - 3x )

Solution--->

Let

y = Cos⁻¹ ( 4x³ - 3x)

Putting x = Cosθ => θ = Cos⁻¹x

y = Cos⁻¹ ( 4 Cos³θ - 3 Cosθ )

We have a formula

Cos 3A = 4 Cos³A - 3CosA ,Applying it here

= Cos⁻¹ ( Cos3θ )

= 3 θ

= 3 Cos⁻¹x

Differentiating with respect to x

dy/dx = 3 d / d x ( Cos⁻¹x )

= 3 { - 1 / √(1 - x² ) }

= - 3 / √(1 - x² )

Additional information

1) Sin2A = 2 SinA CosA

= 2 tanA / 1 + tan²A

2) Cos2A = 2 Cos²A - 1

= 1 - 2 Sin²A

= Cos²A - Sin²A

= 1 - tan²A / 1 + tan²A

3) tan2A = 2 tanA / 1 - tan²A

4) Sin3A = 3 SinA - 4 Sin³A

Answer 2

Answer:

Step-by-step explanation:

Cos⁻¹ ( 4x³ - 3x )

Let

y = Cos⁻¹ ( 4x³ - 3x)

 x = Cosθ

=> θ = Cos⁻¹x

y = Cos⁻¹ ( 4 Cos³θ - 3 Cosθ )

By a formula

Cos 3A = 4 Cos³A - 3CosA ,

= Cos⁻¹ ( Cos3θ )

= 3 θ

= 3 Cos⁻¹x

now by differentiation

dy/dx = 3 d / d x ( Cos⁻¹x )

= 3 { - 1 / √(1 - x² ) }

= - 3 / √(1 - x² )

              ................