Question

Using the chain rule, find the respective derivatives of m(t)=1+(0.01t^1.3) & T(m)=15(m+1)^1.14. Please show steps, I'm trying to learn to do the chain rule!

Answer 1

I'll show you how to find T ' (m). This is the derivative of T(m).



\[\Large T(m) = 15(m+1)^{1.14}\]

\[\Large \frac{d}{dm}[T(m)] = \frac{d}{dm}[15(m+1)^{1.14}]\]

\[\Large T^{\prime}(m) = \frac{d}{dm}[15(m+1)^{1.14}]\]

\[\Large T^{\prime}(m) = 15\frac{d}{dm}[(m+1)^{1.14}]\]

\[\Large T^{\prime}(m) = 15*1.14*(m+1)^{1.14-1}\frac{d}{dm}[m+1]\]

\[\Large T^{\prime}(m) = 17.1*(m+1)^{0.14}*1\]

\[\Large T^{\prime}(m) = 17.1(m+1)^{0.14}\]

Answer 2

I used the coefficient rule, the power rule and the chain rule.

Answer 3

Thank you! If anyone has time I'm really struggling with the first one \:

Answer 4

were you able to get started?

Answer 5

or no?

Answer 6

I gave it a go but the +1 is throwing me off, because it's a constant will it automatically be eliminated?

Answer 7

yeah you break it up and it turns out deriving a constant leads to 0

Answer 8

Okay! I'll keep working on that one then and let you know if I get stuck again (:

Answer 9

ok sounds good. Let me know what you get

Answer 10

dunno if you're still around but I'm having trouble finding a place to start on this \: sorry, was sick for this day in class!

Answer 11

For m(t), you don't need to use the chain rule.


\[\Large m(t)=1+(0.01t^{1.3})\]


\[\Large m(t)=1+0.01t^{1.3}\]


\[\Large \frac{d}{dt}[m(t)]=\frac{d}{dt}[1+0.01t^{1.3}]\]


\[\Large m^{\prime}(t)=\frac{d}{dt}[1+0.01t^{1.3}]\]


\[\Large m^{\prime}(t)=\frac{d}{dt}[1]+\frac{d}{dt}[0.01t^{1.3}]\]


\[\Large m^{\prime}(t)=0+\frac{d}{dt}[0.01t^{1.3}]\]


\[\Large m^{\prime}(t)=\frac{d}{dt}[0.01t^{1.3}]\]


\[\Large m^{\prime}(t)=0.01*\frac{d}{dt}[t^{1.3}]\]


\[\Large m^{\prime}(t)=0.01*1.3*t^{1.3-1}\]


\[\Large m^{\prime}(t)=0.013t^{0.3}\]

Answer 12

ahhh okay! I had gotten that before but assumed I was doing it wrong. Thank you so much!

Answer 13

np