differentiate the ffg wrt x: y = e^(-2x) . sin 5x y = 2cot (x^2 - 3) y = ln [x^2 .square root (1-sinx)]

We'll differentiate the first composed
function:

y = e^(-2x) . sin
5x

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

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

y' = [e^(-2x) . sin
5x]'

y' = [e^(-2x)]'*(sin 5x) + [e^(-2x)]*(sin
5x)'

y' = e^(-2x)*(-2x)'*(sin 5x) + [e^(-2x)]*(cos
5x)*(5x)'

y' = -2e^(-2x)*(sin 5x) + 5*[e^(-2x)]*(cos
5x)

We'll factorize and the final result will
be:

y' = e^(-2x)*(-2sin 5x + 5 cos
5x)

We'll differentiate the
second function y = 2cot (x^2 - 3)

y' = [2cot (x^2 -
3)]'

y' = -2(x^2 - 3)'/[sin (x^2 -
3)]^2

y' = -2*2x/[sin (x^2 -
3)]^2

y' = -4x/[sin (x^2 -
3)]^2

We'll differentiate the
third given function

y = ln [x^2 *square root
(1-sinx)]

y' = {ln [x^2 *square root
(1-sinx)]}'

y' = [x^2 *square root (1-sinx)]'/x^2*square
root (1-sinx)]

Since the numerator is a product, we'll
apply the product rule:

[x^2 *square root (1-sinx)]' =
(x^2)'*sqrt(1-sinx)+(x^2)*(1-sinx)'

[x^2 *square root
(1-sinx)]' = (2x)*sqrt(1-sinx)+(x^2)*(-cos
x)

y' = [(2x)*sqrt(1-sinx)+(x^2)*(-cos
x)]/x^2*square root (1-sinx)