Differentiate the following formula: j(t) = 250 / (1 + 15 * e^(-0.18t))
Understand the Problem
The question asks to find the derivative of the function j(t). This will involve applying the quotient rule and chain rule of differentiation to find j'(t).
Answer
$j'(t) = \frac{675e^{-0.18t}}{(1 + 15e^{-0.18t})^2}$
Answer for screen readers
$$j'(t) = \frac{675e^{-0.18t}}{(1 + 15e^{-0.18t})^2}$$
Steps to Solve
- Rewrite the function
Rewrite $j(t)$ to make differentiation easier.
$$ j(t) = \frac{250}{1 + 15e^{-0.18t}} = 250(1 + 15e^{-0.18t})^{-1} $$
- Apply the chain rule
Differentiate $j(t)$ with respect to $t$ using the chain rule.
$$ j'(t) = 250 \cdot (-1) (1 + 15e^{-0.18t})^{-2} \cdot \frac{d}{dt}(1 + 15e^{-0.18t}) $$
- Differentiate the inner function
Differentiate the inner function $1 + 15e^{-0.18t}$ with respect to $t$.
$$ \frac{d}{dt}(1 + 15e^{-0.18t}) = 0 + 15 \cdot e^{-0.18t} \cdot (-0.18) = -2.7e^{-0.18t} $$
- Substitute back into the expression for $j'(t)$
Substitute the derivative of the inner function back into the expression for $j'(t)$.
$$ j'(t) = -250 (1 + 15e^{-0.18t})^{-2} \cdot (-2.7e^{-0.18t}) $$
- Simplify the expression
Simplify the expression for $j'(t)$.
$$ j'(t) = \frac{675e^{-0.18t}}{(1 + 15e^{-0.18t})^2} $$
$$j'(t) = \frac{675e^{-0.18t}}{(1 + 15e^{-0.18t})^2}$$
More Information
The derivative $j'(t)$ represents the rate of change of the function $j(t)$ with respect to $t$. This tells you how the value of the function is changing as $t$ changes.
Tips
A common mistake is forgetting to apply the chain rule when differentiating the exponential term. Another common mistake is making errors when simplifying the expression after applying the differentiation rules. Be careful with the signs and constants.
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