Differentiation of Simple Functions
Session Preparation¶
Brooks: Chapter 10
Some of the exercises may require you to use Python. You may also need to install the numpy and sympy libraries if you haven't already.
Resources Danish Class:¶
Exercises¶
[Python] Means that you are advised to use Python to solve the exercise. You can find the Python solutions here or download them here
Exercise 1 (Recap)¶
[Python] Given
\(A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{array}\right]\) and \(B=\left[\begin{array}{ccc}9 & 8 & 7 \\ 6 & 5 & 4 \\ 3 & 2 & 1\end{array}\right]\).
Solve the matrix equation \(X-A=6(X+B)\).
\(X=\left[\begin{array}{ccc}-11 & -10 & -9 \\ -8 & -7 & -6 \\ -5 & -4 & -3\end{array}\right]\)
Exercise 2¶
Find the limits of the following functions:
- \(\lim\limits_{x \to 0} \frac{x^4-4 x^3+x^2}{x^3+x^2+x}\)
- \(\lim\limits_{x \to \infty} \frac{x^2+1}{x^2-1}\)
- \(\lim\limits_{x \to 2} \frac{x-2}{x^2-3 x+2}\)
- \(\lim\limits_{x \to 0} \frac{x^4-4 x^3+x^2}{x^3+x^2+x} = 0\)
- \(\lim\limits_{x \to \infty} \frac{x^2+1}{x^2-1} = 1\)
- \(\lim\limits_{x \to 2} \frac{x-2}{x^2-3 x+2} = 1\)
Exercise 3: Simple Derivatives¶
Find the derivatives of the following functions:
- \(f(x)=x^4\)
- \(f(t)=t^{-\frac{1}{3}}\)
- \(y=t^{-3.8}\)
- \(f'(x)=4x^3\)
- \(f'(t)=-\frac{1}{3}t^{-\frac{4}{3}}\)
- \(y'=-3.8t^{-4.8}\)
Exercise 4: Derivatives of Powers¶
Express the following as powers and then differentiate:
- \(\frac{1}{x^2}\)
- \(\sqrt[3]{x}\)
- \(\frac{1}{x \sqrt[4]{x}}\)
- \(\frac{1}{x^2} = x^{-2}\), derivative: \(\frac{d}{dx}x^{-2} = -2x^{-3}\)
- \(\sqrt[3]{x} = x^{1/3}\), derivative: \(\frac{d}{dx}x^{1/3} = \frac{1}{3}x^{-2/3}\)
- \(\frac{1}{x \sqrt[4]{x}} = x^{-5/4}\), derivative: \(\frac{d}{dx}x^{-5/4} = -\frac{5}{4}x^{-9/4}\)
Exercise 5: Derivatives of Sums and Differences¶
Find the derivatives of the following functions:
- \(y=2 x^{-7}+\frac{3}{x^2}\)
- \(f(u)=u^{\frac{5}{3}}-3 u^{-7}\)
- \(g(z)=8 z^{-2}-\frac{5}{z}\)
- \(\frac{d}{dx}(2 x^{-7}+\frac{3}{x^2}) = -14x^{-8} - 6x^{-3}\)
- \(\frac{d}{dx}(u^{\frac{5}{3}}-3 u^{-7}) = \frac{5}{3}u^{2/3} + 21u^{-8}\)
- \(\frac{d}{dx}(8 z^{-2}-\frac{5}{z}) = -16z^{-3} + \frac{5}{z^2}\)
Exercises 6: Derivatives of Products and Quotients¶
Use the product rule to differentiate the functions below:
- \(f(x)=\left(4 x^3+2\right)(1-3 x)\)
- \(g(x)=\left(x^2+x+2\right)\left(x^2+1\right)\)
- \(h(x)=\frac{x^2-1}{x^3+4}\)
- \(g(t)=\frac{t(t+6)}{t^2+3 t+1}\)
- \(-48x^3 + 12x^2 - 6\)
- \( 4x^3 + 3x^2 + 6x + 1\)
- \(h^{\prime}(x)=\frac{-x^4+3 x^2+8 x}{\left(x^3+4\right)^2}\)
- \(\frac{-3 t^2+2 t+6}{\left(t^2+3 t+1\right)^2}\)
Exercise 7: Derivatives using the Chain Rule¶
- \((2 x+3)^2\)
- \(\left(x^2+2 x+1\right)^{12}\)
- \(f(t)=\sqrt{t^2-5 t+7}\)
- \(z=\left(x+\frac{1}{x}\right)^{\frac{3}{7}}\)
- \(\frac{x}{\sqrt{1-x^2}}\)
- \(8 x+12\)
- \(12(2 x+2)\left(x^2+2 x+1\right)^{11}\)
- \(\frac{2 t-5}{2 \sqrt{t^2-5 t+7}}\)
- \(\frac{3}{7}\left(x+\frac{1}{x}\right)^{-4 / 7}\left(1-\frac{1}{x^2}\right)\)
- \(\frac{1}{\left(1-x^2\right)^{3 / 2}}\)
Exercise 8: Derivatives of Exponential and Logarithmic Functions¶
- \(f(x)=\ln \left(2 x^3\right)\)
- \(f(x)=e^{x^2+x^3}\)
- \(f(x)=\ln \left(e^x+x^3\right)\)
- \(\frac{3}{x}\)
- \(e^{x^2+x^3} (2x+3x^2)\)
- \(\frac{e^x+3x^2}{e^x+x^3}\)
Exercise 9: Applications of Derivatives¶
Velocity is the rate of change of position with respect to time.
Acceleration is the rate of change of velocity with respect to time.
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Given the position function of a car \(s(t) = 5t^2 + 2t + 1\), where \(t\) is in seconds and \(s(t)\) is in meters, find the velocity function.
Velocity function: \(v(t) = 10t + 2\) m/s
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Knowing that acceleration is the derivative of velocity, find the acceleration function.
Acceleration function: \(a(t) = 10\) m/s²
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Calculate the velocity and acceleration of the car at \(t = 3\) seconds.
Velocity at \(t = 3\) seconds: \(v(3) = 32\) m/s
Acceleration at \(t = 3\) seconds: \(a(3) = 10\) m/s²