Question

1. Using the Chain Rule, find dw/dt when:
w = 2ye^(x) - ln z; x = ln (t^(2) + 1) ; y = tan t; z = e^(t)

Answer 1

\(w = 2ye^x - \ln z\)

\(w = 2(\tan t) e^{ ln (t^2 + 1)} - \ln (e^t)\)

Answer 2

In general, what is \(e^{\ln x} \) equal to?

Answer 3

x

Answer 4

Also, what is \( \ln e^x\) equal to?

Answer 5

1

Answer 6

No, it's also x.

Answer 7

Using those two substitutions will make your expression much simpler.
Then you can differentitate it.

Answer 8

\(w = 2(\tan t) e^{ ln (t^2 + 1)} - \ln (e^t)\)

First, substitute \(t^2 + 1 \) for \(e^{ ln (t^2 + 1)}\) and \(t \) for \(\ln (e^t)\).

Answer 9

w = (2t^2+2)(tant)-t

Answer 10

Good.
Now remember that the derivative of a sum is the sum of the derivatives.
\(\dfrac{d}{dx} [f(x) + g(x)] = \dfrac{d}{dx} f(x) + \dfrac{d}{dx} g(x) \)

Answer 11

That means:

\( \dfrac{dw}{dt} = \dfrac{d}{dt}[ (2t^2+2)(tant)-t] \)

\( = \dfrac{d}{dt}[ (2t^2+2)(tant)] - \dfrac{d}{dt} t \)

Answer 12

FS'+SF'
(2t^2+2)(sec^2(x))+tan(t)(4t) - 1

Answer 13

Now you need the product rule on the left part, and a simple derivative on the right part.

Answer 14

I see that you are ahead of me.
Exactly. That's correct.

Answer 15

excellent, thank you

Answer 16

Actually, the excellent goes to you! Good job.
You're welcome.