Differentiate y wrt x y=2^-x*cos2x y=tan[sqrt(x^2-2)] .

To differentiate  y with respect to
x

(1) y = 2^-x*cosx (2) y = tan
(x^2-x).

Solution:

(1)

y
= 2^(-x) * cos2x.

We know that y'  = (u*v)'x = u'(x)*v(x)
+u(x)*v'(x).

u(x) = 2^-x. u'(x) =  2^(-x)*
ln2.

v(x) = cos(2). v'(x) = -(sin2x)*(2x)' =
-2sin2x

Therefore ,

y' =
{(2^-x)*cos2x}' = (2^-x)'*cos2x + (2^-x)(cos2x)'

=
(2^-x)(ln2) cos2x+ 2^-x * (-2sin2x)

y' = 2^-x{ ln2 * cos2x
-2sin2x}

(2)

y = tan
{sqrt(x^2-2)}

We know that if y = u (v(x) ,  then y' =
{u'(v(x)}*v'(x)

So

(x^2-2}' =
2x

{sqrt(x^2-2)}' = {(x^2-2)^(1/2)
}'

= (1/2) (x^2-2)^(1/2 -1 ) *
(x^2-2)'

= (1/2) (x^2-2)^(-1/2) *
2x

=  -x/(x^2-2)^(1/2).

tan
(sqrt(x^2-2) = {sec^2 [sqrt(x^2-2)] } {sqrt(x^2-2)}'

=
{[sec(sqrt(x^2-2))]^2} {-x/(x^2-2)^(1/2)}

tan {sqrt(x^2-2)}
= {-x/(x^2-2)} { [ sec (sqrt(x^2-2))]^2}