For #1 notice that .
#1 Find y' of y=9^(-x)
My work:
y' = 9^(-x) * ln(9) * (-x)'
y' = -9^(-x) * ln(9)
#2 Find y' of y = ln((3-x)/(3+x))
y' = 1/((3-x)/(3+x)) * ((3-x)/(3+x))'
y' = 1/((3-x)/(3+x)) * ((3-x)'(3+x)-(3-x)(3+x)')/(3+x)^2
y' = (3+x)/(3-x)* ((-1)(3+x)-(3-x)(1))/(3+x)^2
y' = (3+x)/(3-x)* (-3-x-3+x)/(3+x)^2
y' = 1/(3-x) * -6/(3+x)
y' = -6/((x+3)(x-3)
y' = -6/(x^2-9)
#3 Find y' of y = ln((x(x^2+2)/(sqrt(x^3-7))
Answers from my teacher:
#1 (-2)(3^(-2x)(ln(3))
#2 6/(x^2-9)
#3 (1/x) + (2x)/(x^2+2) - (3x^2)/(2(x^3-7))
At penultimate line of #2 there shouldn't be a minus sign since you change from ) to . Furthermore there is a much easier way of doing #2 by first applying the relationship of logarithms that states that
before differentiation.
Trust me this way takes a lot less work!
Applying the same method to #3 will also give you the required result quite easily.