Derivative Summary

In order to differentiate this expression, we can use Product Rule.

First, we can identify \(f(x)\) and \(g(x)\):

\(\textcolor{red}{f(x) = \sin(x^2)}\)

\(\textcolor{blue}{g(x) = \cos(5^x)}\)

In order to determine \(f'(x)\), we can use Chain Rule:

\(\textcolor{red}{f'(x) = \cos(x^2)(2x)}\)

In order to determine \(g'(x)\), we can use Chain Rule:

\(\textcolor{blue}{g'(x) = -\sin(5^x)(5^x\cdot \text{ln}(5))}\)

We can now use Product Rule to differentiate this expression:

\(y' = \textcolor{red}{f'(x)}\textcolor{blue}{g(x)} + \textcolor{blue}{g'(x)}\textcolor{red}{f(x)}\)

\(y' = \textcolor{red}{\cos(x^2)(2x)}\textcolor{blue}{\cos(5^x)} + [\textcolor{blue}{-\sin(5^x)(5^x\cdot \text{ln}(5))}\textcolor{red}{\cos(x^2)(2x)}]\)

\(y' = \cos(x^2)(2x)\cos(5^x) - \sin(5^x)(5^x\cdot \text{ln}(5))\cos(x^2)(2x)\)

Therefore, we can determine that \(y' = \cos(x^2)(2x)\cos(5^x) -\sin(5^x)(5^x\cdot \text{ln}(5))\cos(x^2)(2x)\).

First, we can determine \(f'(g(x))\). This involves using Chain Rule twice. We can use it to multiply the original function by its exponent (4):

\(= 4[\sin(5x+e^x)]^3\)

We can then use it to multiply the function by its trignometric derivative, \(\cos\):

\(\cos(5x+e^x)\)

When we put these 2 portions together, we get \(f'(g(x))\):

\(\textcolor{red}{f'(g(x)) = 4[\sin(5x+e^x)]^3\cos(5x+e^x)}\)

Next, we can determine \(g'(x)\). This involves using Power Rule and the Natural Exponential Formula:

\(\textcolor{blue}{g'(x) = 5+e^x}\)

Finally, we can use the Chain Rule to find the derivative of the whole expression:

\(y' = \textcolor{red}{f'(g(x))}\cdot \textcolor{blue}{g'(x)}\)

\(y' = \textcolor{red}{4[\sin(5x+e^x)]^3\cos(5x+e^x)}[\textcolor{blue}{5+e^x}]\)

Therefore, we can determine that \(y' = 4[\sin(5x+e^x)]^3\cos(5x+e^x)[5+e^x]\).

First, we can substitute \(\cfrac{e}{2}\) for \(x\) in the original expression to determine its respective \(y\)-coordinate:

\(y \left(\cfrac{e}{2}\right) = \text{ln}(2) \left(\cfrac{e}{2}\right)\)

\(y \left(\cfrac{e}{2}\right) = \text{ln}(e)\)

\(y \left(\cfrac{e}{2}\right) = 1\)

Therefore, we can determine that the Point is \(\left(\cfrac{e}{2}, 1\right)\).

Next, we can differentiate the original expression in order to find the equation of its tangent:

\(y' = \left(\cfrac{1}{2x}\right)(2)\)

\(y' = \cfrac{1}{x}\)

Then, we can substitute \(\cfrac{e}{2}\) for \(x\) to determine the tangent's slope:

\(y\left(\cfrac{e}{2}\right) = \cfrac{1}{e/2}\)

\(m = \cfrac{1}{e/2}\cdot \cfrac{2}{2}\)

\(m = \cfrac{2}{e}\)

Finally, we can substitute all the pertinent values into the slope-intercept equation to determine the \(y\)-intercepts:

\(y = mx + b\)

\(1 = \left(\cfrac{2}{e}\right) \left(\cfrac{e}{2}\right) + b\)

\(1 = 1 + b\)

\(b = 0\)

\(y = \cfrac{2}{e}x\)

Therefore, we can determine that the equation of the tangent is \(y = \cfrac{2}{e}x\).

Given this information, we can sketch a graph of the original expression and its tangent at \(x = \cfrac{e}{2}\):

i. First, we can differentiate the original function in order to determine the velocity function:

\(y' = 1e^t + te^t - e^t - 4t\)

\(y' = te^t - 4t\)

Next, we can set the function equal to \(0\). We can then factor the function to determine its roots:

\(0 = te^t - 4t\)

\(0 = t(e^t - 4)\)

\(t = 0\)

In order to determine the root of \(e^t - 4\), we can set the 2 terms equal to each other. We can then find the \(ln\) of \(4\) to determine solve for that root:

\(e^t - 4 = 0\)

\(e^t = 4\)

\(t = \ln4\)

\(t = 1.39\)

Therefore, we can determine that the velocity equals \(0\) when \(t = 0\;[\text{s}]\) or \(t = 1.39\;[\text{s}]\).

---

ii. In order to determine if there is a maximum or minimum distance to the origin, we can use the First Derivatives Test using the zeroes that we determined in the last question:

Interval	Test \(x\)-value	\(f'(x)\)	Conclusion
\(\textcolor{green}{(-∞, 0)}\)	\(-1\)	\(f'(-1)= 3.63\)	\(f\) is Increasing
\(\textcolor{red}{(0, 1.39)}\)	\(1\)	\(f'(1)= -1.28\)	\(f\) is Decreasing
\(\textcolor{green}{(1.39, ∞)}\)	\(2\)	\(f'(2)= 6.78\)	\(f\) is Increasing

Therefore, we can determine that there is a minimum at \(t = 1.39\;[\text{s}]\).

---

iii. In order to determine the acceleration, we can differentiate the velocity function:

\(f'' = 1e^t + te^t - 4\)

---

iv. In order to determine if there is a max/min velocity, we can factor the acceleration function to help determine its zeroes:

\(f'' = e^t(1 + t - 4e^{-t})\)

As you can see, when we factored \(e^t\) out of the function, we multiplied \(4\) by \(e^{-t}\). This is because these values will end up cancelling each other out.

As \(e^t\) will never equal \(0\), we can determine if \(1 + t\) will intersect with the exponential (\(4e^{-t}\)) to determine if there is a max/min. We can determine this more easily by creating a table of values:

\(x\)-value	-1	0	1	2	3
1 + t	0	1	2	3	4
4e⁻ᵗ	10.87	4	1.47	0.54	0.199

Based on the table of values, we can determine that the functions intersect somewhere between \(x = 0\) and \(x = 1\). We can sketch a graph to get a visual indication of where these functions intersect:

Therefore, we can determine that there is a max/min value around \(x = 0.8\).

i. First, we can determine the derivative of the original equation:

\(C'(t) = 150(0.5)^t + 150t(0.5)^t\cdot \ln(0.5)\)

Next, we can factor this equation in order to determine its zeroes more easily:

\(C'(t) = 150(0.5)^t[1 + t\cdot \ln(0.5)]\)

We can determine that since \(150\) and \((0.5)^t\) can never equal \(0\) that they can't count as Critical Points. As a result, we can determine the Critical Point of \(1 + t\cdot \ln(0.5)\):

\(0 = 1 + t\cdot \ln(0.5)\)

\(t\cdot \ln(0.5) = -1\)

\(\cfrac{t\cdot \cancel{\ln(0.5)}}{\cancel{\ln(0.5)}} = \cfrac{-1}{\ln(0.5)}\)

\(t = 1.44\)

We can now use the First Derivatives Test to determine the highest concentration of the medicine:

Interval	Test \(x\)-value	\(f'(x)\)	Conclusion
\(\textcolor{green}{(-∞, 1.44)}\)	\(1\)	\(f'(1)= 23\)	\(f\) is Increasing
\(\textcolor{red}{(1.44, ∞)}\)	\(1.5\)	\(f'(1.5)= -2\)	\(f\) is Decreasing

We can determine that there is a maximum at \(t = 1.44\;[s]\).

We can substitute \(1.44\) for \(t\) in the original function to determine its concentration value at that time:

\(C(1.44) = 150(1.44)(0.5)^{1.44}\)

\(C(1.44) = 216(0.37)\)

\(C(1.44) = 79.61\)

Therefore, we can determine that the highest concentration value of the medicine is \(79.61\;[\text{parts}/\text{million}]\).

---

ii. In order to determine how quickly the concentration is decreasing after \(2\) hours, we can substitute \(2\) into the derivative function:

\(C'(2) = 150(0.5)^2 + 150(2)(0.5)^2\cdot \ln(0.5)\)

\(C'(2) = 150(0.25) + 300(0.25)(-0.69)\)

\(C'(2) = 37.5 - 51.75\)

\(C'(2) = -14.25\)

Therefore, we can determine at \(2\) hours, the medicine is decreasing at a rate of \(-14.25\; [\text{parts per milllion}/\text{hour}]\).