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)