Derivative? f(x) = ln(3-e^(-x))

[JSS6151]

we're learning about the product rule and i understand it in class but I'm having a really hard time not making mistakes in my work, plus the ln is throwing me off...my homework is for accuracy too PLEASE help!

 

[SammyS]

This problem requires using the chain rule, not the product rile.

What have you tried?

Where are you stuck?

 

[Mark44]

we're learning about the product rule and i understand it in class but I'm having a really hard time not making mistakes in my work, plus the ln is throwing me off...my homework is for accuracy too PLEASE help!

The product rule doesn't apply in this problem, but the chain rule does.

 

[cepheid]

Welcome to PF!

The chain rule will be necessary here. If it helps, think of it as doing multiple substitutions so that:

u(x) = 3-e^-x

f(u) = ln(u)

Then the chain rule says:

$$\frac{df}{dx} = \frac{df}{du}\frac{du}{dx}$$

 

[JSS6151]

my mistake. we're learning about chain, product, and quotient rule all at once so i got confused.

our professor was teaching us to separate the problem into two pieces, so i know to make it something like

f(x) = ln(z) and f(z)=3-e^(-x))

then the derivative of f(x) would be 1/z and you'd plug in f(z) into that but i don't know the derivative of f(z) and I'm not even sure if what I've posted so far is correct!

 

[Mark44]

Don't use the same letter for two different functions. That's a surefire way to get confused. And don't write f(x) when your function actually involves z, and vice-versa.

f(z) = ln(z)
g(x) = 3 - e^-x

 

[cepheid]

f(x) = ln(z) and f(z)=3-e^(-x))

Your notation here doesn't really make sense. If something is a function of x, then it should depend on x. If something is a function of z, then it should depend on z. Also, you should use different symbols for the "inside" and "outside" functions. To see how to express the composition of functions properly, see my post above.

As for how to differentiate 3-e^-x, just apply the rules. The derivative of a sum is the sum of the derivatives, so that:

$$\frac{d}{dx}(3-e^{-x}) = \frac{d}{dx}(3) - \frac{d}{dx} (e^{-x})$$

Now, to differentiate the exponential, you can apply the chain rule once again:

e^-x = e^g(x)

where g(x) = -1x

and the chain rule says:


$$\frac{d}{dx}(e^{g(x)}) = \frac{d}{dg}(e^{g(x)})\frac{dg(x)}{dx}$$

 

[JSS6151]

thanks for the advice
so if f(z) = lz(z) and g(x)=3-e^-x

f'(z) = 1/z right? because derivative of ln x = 1/x.

and g'(x) = 3-(-1)e^x because derivative of e^x = e^x

not sure about that last line i wrote...
and then i would plug the answer for g'(x) into the 1/z in place of z?

 

[Mark44]

thanks for the advice
so if f(z) = lz(z) and g(x)=3-e^-x

f'(z) = 1/z right? because derivative of ln x = 1/x.

Ignoring the typo "lz(z)", yes.

and g'(x) = 3-(-1)e^x because derivative of e^x = e^x

No. The derivative of a constant is 0, so you should get g'(x) = +e^-x. Note the minus sign on the exponent.

not sure about that last line i wrote...
and then i would plug the answer for g'(x) into the 1/z in place of z?

You have f(z) = ln(z), and u = g(x) = 3 - e^-x
For d/dx(f(g(x)) = d/dx(f(u)), you should get f'(u)*u'(x)