# Math Help - Derivative Of Inverse Trigonometry, Exponents, And Log

1. ## Derivative Of Inverse Trigonometry, Exponents, And Log

#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))

#1 (-2)(3^(-2x)(ln(3))
#2 6/(x^2-9)
#3 (1/x) + (2x)/(x^2+2) - (3x^2)/(2(x^3-7))

2. For #1 notice that $9^{-x}=3^{-2x}$.

3. Originally Posted by Plato
For #1 notice that $9^{-x}=3^{-2x}$.
Big help got #1 now

4. At penultimate line of #2 there shouldn't be a minus sign since you change from $(3-x$) to $-(x-3)$. Furthermore there is a much easier way of doing #2 by first applying the relationship of logarithms that states that

$\log_a \left( \frac{x}{y} \right) = \log_a x - \log_a y$

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.