Question

Find the dy/dx

y= sin²x/cos²x​

Answer 1

Since we have

`y = (\sin^2(x))/(\cos^2(x)) = \tan^2(x)`

by definition of tangent, differentiating with the chain rule gives

`(dy)/(dx) = \boxed{2 \tan(x) \sec^2(x)}`

(second choice)

Answer 2

d y dx=0

Step-by-step explanation:

From the given

y=sin2x+cos2x

The right side of the equation is

=1

y=1

d y d x=d d x(1)=0

or we can do it this way.

y = sin2x+cos2x

d y d x=2

⋅(sinx)2−1

⋅

d d x(sinx)+2

(cosx)2−1 d d x(cosx)

d y d x=2⋅

sinx⋅cosx+2⋅cosx⋅(−sinx)

dydx=2⋅sinx⋅cosx−2⋅sinx⋅cosx

dydx=0

God bless....I hope the explanation is useful.