Hi

I am struggling to solve the equations:

Given:

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

use the chain rule to find dy/dx:

a) y=e^5t,

b) y=4e^3t

c)y=6e^-2t

(where "^" represents an exponent)

Thanks in advance for any help

Motty

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- May 20th 2009, 01:38 AMmottyusing chain rule to solve an equation
Hi

I am struggling to solve the equations:

Given:

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

use the chain rule to find dy/dx:

a) y=e^5t,

b) y=4e^3t

c)y=6e^-2t

(where "^" represents an exponent)

Thanks in advance for any help

Motty - May 20th 2009, 03:11 AMSpec
$\displaystyle y=e^{5t}=\{u=5t\}=e^{u(t)}$

$\displaystyle \frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}=e^u\frac{ du}{dt} = e^{u}\cdot 5 = 5e^{5t}$

Use the same process for the other questions. - May 20th 2009, 03:11 AMScott H
The Chain Rule states,

$\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))g'(x).$

It can be seen to work through the fact that $\displaystyle g(x)$ changes at the rate $\displaystyle g'(x)$ as $\displaystyle x$ moves, speeding up (or slowing down) the rate of change of $\displaystyle f(g(x))$ by that amount.

For instance, if $\displaystyle g'(x)=2$, then $\displaystyle g(x)$ moves twice as fast, and $\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot 2$.

For problem (a), we note that $\displaystyle f(t)=e^t$ and $\displaystyle g(t)=5t$. The Chain Rule therefore states,

$\displaystyle \frac{d}{dt}(e^{5t})=\frac{d}{dt}f(g(t))=f'(g(t))g '(t)=e^{g(t)}\frac{d}{dt}(5t)=e^{5t}\cdot 5=5e^{5t}.$

The same reasoning may be applied to (b) and (c). - May 20th 2009, 04:54 AMmotty
Thanks guys, following the same reasonig, I think, does that mean:

b) = 12e^3t

c) = -12e^-2t