Question

if y=(cos x) ^ x find dy/dx​

Answer 1

Answer:

y = (cos x)^x

Taking natural log on both sides,

ln y = x ln cos x

Now, differentiating both sides wrt x

We get,

1/y * dy/dx = x (-sinx/cosx) + ln cos x * (1)

So,

dy/dx = (cosx)^x { ln cos x - x tan x }

Answer 2

Answer:

( cos x )^x ( ㏑ cos x - x . ( tan x ) )

Step-by-step explanation:

Given :

y = ( cos x )^x

Taking ㏑ both side we get :

= > ㏑ y = ㏑ ( cos x )^x

= > ㏑ y = x . ㏑ cos x

Diff. w.r.t. x :

= > 1 / y ( d y / d x ) = x ( ㏑ cos x )' + ㏑ cos x ( x )'

= >  1 / y ( d y / d x ) = x ( ㏑ cos x )' + ㏑ cos x . 1

= >  1 / y ( d y / d x ) = x ( ㏑ cos x )' + ㏑ cos x

= >   1 / y ( d y / d x ) = x /  cos x . ( cos x )' + ㏑ cos x

= >  1 / y ( d y / d x ) = x /  cos x . ( - sin x ) + ㏑ cos x

= >  1 / y ( d y / d x ) = - x . ( sin x /  cos x ) + ㏑ cos x

Putting value of y = ( cos x )^x

= > d y / d x = ( cos x )^x ( ㏑ cos x - x . ( tan x ) )

Hence we get required answer!