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

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y= 2^-x * cos2x

we need to diffirentiate y:

y= u*v such that:

u= 2^-x ==> u' = -(2^-x) *ln 2

v= cos2x ==> v = -2sin2x

Then y'= u'v + uv'

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

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

** **

y=tan[sqrt(x^2-2)] .

let u = x^2 - 2 ==> du = 2x dx

==> y= tan(sqrt(u)

y' = (tan(sqrt(u)))'

= u' sqrtu* (tan)' du

= 2x * sqrt(u)* [cos(sqrtu)]^2 *du/2

= x*sqrt(x^2-2)*[cos(sqrt(x^2-2)^2

We'll differentiate the first composed function:

y = 2^(-x)*cos2x

Because it is a product of 2 functions, we'll differentiate according to the product rule:

(u*v)' = u'*v + u*v'

y' = [2^(-x)*cos2x]'

y' = [2^(-x)]'*(cos2x) + [2^(-x)]*(cos2x)'

y' = 2^(-x)*ln 2*(-x)'*(cos2x) + [2^(-x)]*(-sin 2x)*(2x)'

y' = -2^(-x)*ln 2*(cos2x)- 2*[2^(-x)]*(sin 2x)

We'll factorize by -2^(-x) and the final result will be:

**y' = -2^(-x)*[ln 2*(cos2x) + (sin 2x)]**

** **

We'll differentiate the second function y = tan[sqrt(x^2-2)].

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

y' = (x^2-2)'/2*sqrt(x^2-2)*[cos sqrt(x^2-2)]^2

y' = 2x/2*sqrt(x^2-2)*[cos sqrt(x^2-2)]^2

We'll eliminate like terms:

y' = x / sqrt(x^2-2)*[cos sqrt(x^2-2)]^2

**y' = x / sqrt(x^2-2)*[cos sqrt(x^2-2)]^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}