How do you use the chain rule to differentiate #2^(-9z^2+3z+5)#?

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Answer 1 · 2018-04-03T12:56:23

# d/(dz) 2^(-9z^2+3z+5) = (-18z+3)(ln 2) 2^(-9z^2+3z+5) #

We seek:

# dy/(dz)# where #y = 2^(-9z^2+3z+5) #

We can use logarithmic differentiation:

# ln y = ln (2^(-9z^2+3z+5) )#
# \ \ \ \ \ = (-9z^2+3z+5)ln 2 #

Now differentiate Implicitly:

# 1/y dy/(dz) = (-18z+3)ln 2 #

So that:

# dy/(dz) = (-18z+3)(ln 2) y#
# \ \ \ \ \ = (-18z+3)(ln 2) 2^(-9z^2+3z+5) #

We can also use the chain rule directly, and write the function as:

# y = 2^(-9z^2+3z+5) #
# \ \ = e^((ln2)(-9z^2+3z+5)) #

Then applying the chain rule:

# dy/(dz) = e^((ln2)(-9z^2+3z+5)) d/(dz){ (ln2)(-9z^2+3z+5)} #

Giving the same result.