Cant figure out these derivatives

There are a few problems I cant figure out for some reason or another.

they are:

a.) d/dx=(e^x/sqrt(1-x))

b.) d/dx=(swrt(sin(e^x))-9)

c.) d/dx=(x^2ln(cosx))

First, you seem to have problems with notation.

You never write d/dx by itself
It's either dy/dx or d/dx (f(x)), not d/dx = f(x)

Now if we had dy/dx = e^x/sqrt(1-x), this would imply that you would want to find y, using integration. But since title mentions derivatives, then I think you are looking for:

d/dx (e^x / sqrt(1-x)) = ???

See the difference?

Anyway, back to problems. In these problems, you need to use either product rule, quotient rule or chain rule (or a combination of these).

Product rule : d/dx (f(x) g(x)) = f'(x) g(x) + f(x) g'(x)
Quotient rule : d/dx (f(x)/g(x)) = [f'(x) g(x) - f(x) g'(x)] / [g(x)]²
Chain rule : d/dx (f(g(x))) = f'(g(x)) * g'(x)

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a) Use quotient rule and power/chain rule:

d/dx (e^x/√(1-x))
= d/dx (e^x / (1-x)^½)
= [d/dx (e^x) * (1-x)^½ - e^x * d/dx ((1-x)^½)] / ((1-x)½)²
= [e^x (1-x)^½ - e^x * 1/2 (1-x)^-½ * -1] / (1 - x)
= [e^x (1-x)^½ + 1/2 e^x (1-x)^-½] / (1 - x)
= 1/2 e^x (1-x)^-½ [2(1 - x) + 1] / (1 - x)
= e^x (2 - 2x + 1) / (2 (1-x)^½ (1-x))
= e^x (3 - 2x) / (2 (1 - x)^(3/2))

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b) Use chain rule:

d/dx (√(sin(e^x))-9)
= d/dx ((sin(e^x))^(1/2) - 9)
= 1/2 (sin(e^x))^(-1/2) * d/dx sin(e^x) - 0
= 1/2 (sin(e^x))^(-1/2) * cos(e^x) * d/dx (e^x)
= 1/2 (sin(e^x))^(-1/2) * cos(e^x) * e^x
= e^x cos(e^x) / (2 √(sin(e^x)))

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c) Use product rule and chain rule:

d/dx (x² ln(cos x))
= d/dx (x²) * ln(cos x) + x² * d/dx (ln(cos x))
= 2x ln(cos x) + x² (1/cos x) * d/dx (cos x)
= 2x ln(cos x) - x² sin x / cos x
= 2x ln(cos x) - x² tan x

a) quotient rule.
bottom times deriv of top minus the top times the deriv of the bottom all over the bottom squared.

[(1-x)^1/2(e^x)-e^x(1/2(1-x)^-1/2)(-1)]…
from there you can factor out an e^x if you wish or leave it as is.

there is use of the chain rule as well for the derivative of (1-x)^1/2. basically the rule says take the derivative of the outside leaving the inside intact. then multiply by the derivative of the inside.

b)Same goes for this one. chain rule
1/2(sine^x -9)^(-1/2)[cose^x(e^x)]
deriv of out............ deriv of in..

c) uses the product rule and a few laws for logarithms
product rule is first times deriv of second plus second times deriv of first
first being x^2... second being lncosx
x^2[(1/cosx)(-sinx)]+(lncosx)(2x)
1st... der2nd........... 2nd der1st