# Thread: Derivative of Exponential Function Questions

1. ## Derivative of Exponential Function Questions

"For the following, determine the equation of the tangent at the given point:
12b) For the curve defined by (x^2)(e^y)=1 at B(1,0)."

"The number, N, of bacteria in a culture at time t in hours is:
N = 1000[30 + (e^-t/30)]. Find the rate of change of the number of bacteria at time t."

Two different questions, two question I can't solve. Any insight would be appreciated, thanks!

2. Originally Posted by Jeavus
"For the following, determine the equation of the tangent at the given point:
12b) For the curve defined by (x^2)(e^y)=1 at B(1,0)."

"The number, N, of bacteria in a culture at time t in hours is:
N = 1000[30 + (e^-t/30)]. Find the rate of change of the number of bacteria at time t."

Two different questions, two question I can't solve. Any insight would be appreciated, thanks!
For the first one, you need to use Implicit Differentiation. You can start by reshaping the equation so it reads
$x^2=e^{-y}$ When you differentiate both sides, you should get $2x\delta x=-e^{-y} \delta y$

Again, you can reorder this one more time so you have $\frac{\delta y}{\delta x}=[-2xe^y]$

At this point, plug in the point you're given so you get $\frac{\delta y}{\delta x}=-2$

Now you can find the equation for the line using the slope and point you have.

$(y-0)=-2(x-1)$

For the second problem, just use Differentiation to get $\frac{\delta N}{\delta x}$

3. Originally Posted by Jeavus
"For the following, determine the equation of the tangent at the given point:
12b) For the curve defined by (x^2)(e^y)=1 at B(1,0)."

"The number, N, of bacteria in a culture at time t in hours is:
N = 1000[30 + (e^-t/30)]. Find the rate of change of the number of bacteria at time t."

Two different questions, two question I can't solve. Any insight would be appreciated, thanks!
for the first one isolate y first

$x^2e^y=1 \iff e^y=\frac{1}{x^2} \iff y=\ln\left( \frac{1}{x^2}\right) \iff y =\ln(x^{-2}) \iff y=-2\ln(x)$

Now taking the derivative we get

$\frac{dy}{dx}=\frac{-2}{x}$ eval at x=1 and we get

$\frac{dy}{dx}|_{x=1}=\frac{-2}{1}=-2$

Then we get

$y-0=-2(x-1) \iff y=-2x+2$

$
N = 1000[30 + (\frac{e^{-t}}{30})]=3000 + \frac{100}{3}e^{-t}.
$

taking the derivative(is the rate of change) we get

$\frac{dN}{dt}=-\frac{100}{3}e^{-t}$

4. Originally Posted by TheEmptySet
for the first one isolate y first

$x^2e^y=1 \iff e^y=\frac{1}{x^2} \iff y=\ln\left( \frac{1}{x^2}\right) \iff y =\ln(x^{-2}) \iff y=-2\ln(x)$

Now taking the derivative we get

$\frac{dy}{dx}=\frac{-2}{x}$ eval at x=1 and we get

$\frac{dy}{dx}|_{x=1}=\frac{-2}{1}=-2$

Then we get

$y-0=-2(x-1) \iff y=-2x+2$

$
N = 1000[30 + (\frac{e^{-t}}{30})]=3000 + \frac{100}{3}e^{-t}.
$

taking the derivative(is the rate of change) we get

$\frac{dN}{dt}=-\frac{100}{3}e^{-t}$
Haha...or you can just look at the answer that he gave you. lol