Question

y=cos(sinx^2),then at x=√π/2,dy/dx =​

Answer 1

Answer:

√(π/2)sin(1/√2)

Explanation:

Answer 2

Hope u get it .

dy/dx =?

y = cos (sinx²)

Y'= d[cos(sinx²)]/DX

=d[cos(simx²)]/d(sinx²)*d(sinx²)/d(x²)*dx²/dx

= -sin(sin x²)*(cosx²)*(2x)

Putting the value of x as given in the question

-sin(sin π/2).cosπ/2. 2x

As cosπ/2 =0

So any no*0=0

dy/dx or y' = 0