Question

Y = Cos^-1 (log2x) find dy/dx​

Answer 1

Answer:

dy/dx = $\mathsf{\dfrac{- 1} {x{\sqrt{(1 - x^2)} }}}$

Explanation :

dy/dx = $\mathsf{\dfrac{- 2} {2x{\sqrt{( 1 - x^2)} }}}$

dy/dx = $\mathsf{\dfrac{- 1} {x{\sqrt{(1 - x^2)}}}}$

Derivative of cos^-1(x) is $\mathsf{\dfrac{- 1}{{\sqrt{(1 - x^2) }}}}$

Derivative of log2x ( by chain rule) will be 2/2x = 1/x.

Answer 2

Answer:y/dx =  \mathsf{\dfrac{- 1} {x{\sqrt{(1 - x^2)} }}}  

Explanation :

dy/dx =  \mathsf{\dfrac{- 2} {2x{\sqrt{( 1 - x^2)} }}}  

dy/dx =  \mathsf{\dfrac{- 1} {x{\sqrt{(1 - x^2)}}}}  

Derivative of cos^-1(x) is \mathsf{\dfrac{- 1}{{\sqrt{(1 - x^2) }}}}  

Derivative of log2x ( by chain rule) will be 2/2x = 1/x.

Step-by-step explanation: