Question

ahoy !!

can anyone differentiate this?

cos ( 3x - 5 )^2

Answer 1

Hi,

Here is your answer,

y = (cos(3x-5)²

→ dy/dx = 2 cos (3x-5) × -sin(3x-5) × (3(1) - 0)

→ -6 sin (3x-5) cos (3x-5)

NOTE:- We can also elaborate it further.

→ -3 [ 2 sin(3x-5) × cos(3x-5) ]        Let (3x-5) and (3x-5) be θ.

→ dy/dx = -3 × sin2(3x-5)

→  dy/dx = -3sin(6x-10)    {  ∴  FINAL ANSWER   }

Answer 2

$\bf HELLO \: \: DEAR, \\ \\ \bf let \: \: y = cos(3x - 5)^{2} \\ \\ \boxed{\bf put \: \: \: (3x - 5) = t \: \:, y = cost^{2} } \\ \\ \boxed{\bf put \: \: \: {t}^{2} = z \: \: , y = cosz} \\ \\ \bf now \: (3x - 5) = t \: \: , \: \: t^{2} = z \: \: , \: \: y = cosz \\ \\ \bf using \: \: chain \: \: rule \: \: \: dy/dx = dy/dz * dz/dt * dt/dx \\ \\ \bf dy/dx= d(cosz)/dz * d(t^{2})/dt * d(3x - 5)/dx \\ \\ \bf dy/dx = (-sinz) * (2t) * (3) \\ \\ \bf dy/dx = -[sin(3x - 5)^{2} * 2(3x - 5) * (3)] \\ \\ \bf dy/dx = -6(3x - 5) * sin(3x - 5)^{2} \\ \\ \\ \boxed{\underline { \bf \: I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}$