show steps for finding the derivative of y(t) = e^4t?

[Dave411]

Could someone show the steps for finding the derivative of y(t) = e^4t when using the chain rule. In particular, I need to see the steps for case 2 below. This is not a homework problem. It is for my own understanding.

I give 3 ways for solving this below and two of them make sense to me and give the correct answer. However, one case doesn't seem to work.

We start with y(t) = e^4t and the goal is to differentiate it 3 different ways using the chain rule.

To start, I rewrite the y(t) = e^4t in three different ways:

y(t) = (e^4)^t
y(t) = (e^t)^4
y(t) = e^u where u=4t

Here are my steps in each case:

case 1: y(t) = (e^t)^4
f(t) = (x)^4
g(t) = e^t

y'(t) = f'(g(x))*g'(x) (chain rule)
= 4(e^t)^3 * e^t
= 4(e^t)^(3+1)
= 4(e^t)^4
= 4e^(4t) (correct)

case 2: y(t) = (e^4)^t

f(t) = (x)^t
g(t) = e^4

y'(t) = f'(g(x))*g'(x) (chain rule)
= t(e^4)^(t-1) * 0 (e^4 is a constant)
= 0 ???
what went wrong?

case 3:

let u = 4t

d/dt[e^u] = e^u * du/dt
= e^u * 4
= 4e^(4t) (correct)

What is wrong with case 2?

 

[DrPhil]

Could someone show the steps for finding the derivative of y(t) = e^4t when using the chain rule. In particular, I need to see the steps for case 2 below. This is not a homework problem. It is for my own understanding.

I give 3 ways for solving this below and two of them make sense to me and give the correct answer. However, one case doesn't seem to work.

We start with y(t) = e^4t and the goal is to differentiate it 3 different ways using the chain rule.

To start, I rewrite the y(t) = e^4t in three different ways:

y(t) = (e^4)^t
y(t) = (e^t)^4
y(t) = e^u where u=4t

Here are my steps in each case:

case 2: y(t) = (e^4)^t

f(t) = (x)^t
g(t) = e^4.....why not just say x = e^4 = constant = a?

y'(t) = f'(g(x))*g'(x) (chain rule)
= t(e^4)^(t-1) * 0 (e^4 is a constant)....NOT a power rule!
= 0 ???
what went wrong?

Whatr seems to be wrong is how you differentiated f(t) = a^t.
There is no chain, since "t" is the only variable, and it occurs just as itself.
d/dt(a^t) = a^t ln(a)
then backsubstitute a = e^4, ln(a) = 4

 

[Dave411]

d/dt (c^t) $\displaystyle \neq$ t c^t-1 is what went wrong.

c^t = e^ln(c)t

d/dt(e^ln(c))t = ln(c) e^ln(c)t = ln(c) c^t

Thanks. I see that mistake now. But I am still confused.

I rewrite the y(t) = e^4t as y(t) = (e^4)^t
Then (just because I want to understand it) I used the chain rule like this: y'(t) = f'(g(x))*g'(x)

f= h^t and g=e^4

so g' is still going to be zero (because e^4 is a constant) and the chain rule doesn't seem to work. In contrast it seems to work when I define y(t) = (e^t)^4.

 

[Dave411]

There is no chain, since "t" is the only variable, and it occurs just as itself.

OK, so I don't understant the chain rule well enough. But I did read that the power rule is just a specific case of the chain rule... I'm trying to understand all this stuff better.

Why does my case 1 seem to work when using the chain rule?
And why does my cases 3 seem to work when using the power rule (which is a specific case of the chain rule)?

Thanks

 

[pka]

OK, so I don't understand the chain rule well enough.

Well that is your problem.
Why do you expect to understand a question if you lack the basic principles?

 

[JeffM]

Thanks. I see that mistake now. But I am still confused.

I rewrite the y(t) = e^4t as y(t) = (e^4)^t
Then (just because I want to understand it) I used the chain rule like this: y'(t) = f'(g(x))*g'(x)

f= h^t and g=e^4

so g' is still going to be zero (because e^4 is a constant) and the chain rule doesn't seem to work. In contrast it seems to work when I define y(t) = (e^t)^4.

You seem to be asking many questions at once.

Let's make sure you understand $\displaystyle \dfrac{dy}{dx}\ when\ y = r^x,\ where\ r > 0.$

$\displaystyle v = ln(y) = ln\left(r^x\right) = xln(r) \implies \dfrac{dv}{dx} = ln(r).$ With me on that?

$\displaystyle \dfrac{dv}{dx} = \dfrac{dv}{dy} * \dfrac{dy}{dx}.$ Basic chain rule.

$\displaystyle So\ \dfrac{dy}{dx} = \dfrac{dv}{dx} \div \dfrac{dv}{dy}.$ Simple algebra. But what is $\displaystyle \dfrac{dv}{dy}.$

$\displaystyle v = ln(y) \implies \dfrac{dv}{dy} = \dfrac{1}{y} = \dfrac{1}{r^x}.$

$\displaystyle THUS\ \dfrac{dy}{dx} = \dfrac{dv}{dx} \div \dfrac{dv}{dy} = ln(r) \div \dfrac{1}{r^x} = ln(r) * r^x.$

In short, $\displaystyle y = r^x \implies \dfrac{dy}{dx} = ln(r) * r^x.$ In this case, x is a power.

$\displaystyle y = x^r \implies \dfrac{dy}{dx} = rx^{(r - 1)}.$ In this case, x is being raised to a power.

If the power rule shown above can be derived from the chain rule, it would not surprise me, but I do not know that derivation.

Now consider the special case where $\displaystyle y = e^u.$

$\displaystyle y = e^ u \implies \dfrac{dy}{du} = ln(e) * e^u = 1 * e^u = e^u.$

Now if u = f(x)

$\displaystyle y = e^u \implies \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = e^u * \dfrac{du}{dx}.$ Chain rule again.

I hope this clears up some of your confusion.

 

[Dave411]

I hope this clears up some of your confusion.

Thank you!!! That was a great explanation!