Recall, that \(F'(x) \, \, = \, \, f'( g(x) ) \cdot g'(x)\).

\({\bf{a)}}\) \(300 x^2 (x^3-1)^{99}\) .

\({\bf{b)}}\) \(2 (2x+1)^4 (x^3-x+1)^3 (17 x^3 + 6 x^2 - 9x + 3)\).

\({\bf{a)}}\) With \(f(u) = u^3\) and \(g(x) = x^3 - 1\) we have \[ \frac{\textrm{d}}{\textrm{d} \, x} \, (x^3 - 1)^{100} \, \, = \, \, 100 (x^3 - 1)^{99} \cdot 3 x^2 \, \, = \, \, 300 x^2 (x^3-1)^{99} \, . \] \({\bf{b)}}\) Here we must use the Product Rule before using the Chain Rule: \[\begin{eqnarray*} y' & = & \frac{\textrm{d}}{\textrm{d} \, x} \, (2x+1)^5 (x^3-x+1)^4 \\ & = & 5 (2x+1)^4 \cdot 2 \cdot (x^3-x+1)^4 + (2x+1)^5 \cdot 4 (x^3-x+1)^3 \cdot (3y^2-1) \\ & = & 2 (2x+1)^4 (x^3-x+1)^3 (17 x^3 + 6 x^2 - 9x + 3) \, . \end{eqnarray*}\]

Use the chain rule for the both sides of the equation and then collect the terms that involve \(y'\).

\(y' \, \, = \, \, \frac{y^2 \sin(x) + \cos(x+y)}{2y \cos(x) - \cos(x+y)} \,\).

Differentiating implicitly with respect to \(x\) and remembering that \(y = y(x)\) is a function of \(x\), we get \[ \cos(x+y) \cdot (1 + y') \, \, = \, \, 2y \cdot y' \cdot \cos(x) \, + \, y^2 \cdot (-\sin(x)) \, . \] (Note that we have used the Chain Rule on the left side and the Product Rule and Chain Rule on the right side.) If we collect the terms that involve \(y'\), we get \[ \cos(x + y) \, + \, y^2 \sin(x) \, \, = \, \, (2y \cos(x)) \cdot y' \, - \, \cos(x+y) \cdot y' \, . \] Thus, \[ y' \, \, = \, \, \frac{y^2 \sin(x) + \cos(x+y)}{2y \cos(x) - \cos(x+y)} \, . \]

Differentiate the equation with respect to \(x\), find \(y'\) and differentiate the equation again with respect to \(x\) in order to find \(y''\).

\(\, \, -48 \frac{x^2}{y^7} \,\).

Differentiating the equation implicitly with respect to \(x\), we get \[ 4 x^3 \, + \, 4 y^3 y' \, \, = \, \, 0 \qquad \Longrightarrow \qquad y' \, \, = \, \, -\frac{x^3}{y^3} \, . \hspace{3cm} \phantom{u} \] To find \(y''\) we differentiate this expression for \(y'\) using the Quotient Rule and remembering that \(y\) is a function of \(x\): \[\begin{eqnarray*} y'' & = & \frac{\textrm{d}}{\textrm{d} \, x} \left( -\frac{x^3}{y^3} \right) \, \, = \, \, -\frac{3 x^2 \cdot y^3 - x^3 \cdot 3y^2 \cdot y'}{y^6}\\[1mm] & = & -\frac{3 x^2 \cdot y^3 - x^3 \cdot 3y^2 \cdot \left( -\frac{x^3}{y^3} \right)}{y^6} \, \, = \, \, - \frac{3x^2 (x^4 + y^4)}{y^7} \, \end{eqnarray*} \] The values of \(x\) and \(y\) must satisfy the original equation \(x^4 + y^4 = 16\). So the answer simplifies to \[ y'' \, \, = \, \, -\frac{3 x^2 \cdot 16}{y^7} \, \, = \, \, -48 \frac{x^2}{y^7} \, . \]

Keep in mind, that in some cases you might have to use l'Hospital's Rule more than once.

\(\lim_{x \to \infty} \, \frac{{\rm{e}}^x}{x^2} \, = \, \, \infty \, .\)

We have \(\lim_{x \to \infty} \, {\rm{e}}^x = \infty\) and \(\lim_{x \to \infty} \, x^2 = \infty\), so the limit is an indeterminate form of type \(\frac{\infty}{\infty}\), and l'Hospital's Rule (\(\text{l'H}\)) gives \[ \lim_{x \to \infty} \, \frac{{\rm{e}}^x}{x^2} \, \, \stackrel{\text{l'H}}{=} \, \, \lim_{x \to \infty} \, \frac{\, \, \frac{\textrm{d}}{\textrm{d} x} \, {\rm{e}}^x \, \,}{\frac{\textrm{d}}{\textrm{d} x} \, x^2} \, \, = \, \, \lim_{x \to \infty} \, \frac{{\rm{e}}^x}{2x} \] Since \(\lim_{x \to \infty} \, {\rm{e}}^x = \infty\) and \(\lim_{x \to \infty} \, 2x = \infty\), the limit on the right hand side is also an indeterminate form of type \(\frac{\infty}{\infty}\).

A second application of l'Hospital's Rule (\(\text{l'H}\)) gives \[ \lim_{x \to \infty} \, \frac{{\rm{e}}^x}{x^2} \, \, = \, \, \lim_{x \to \infty} \, \frac{{\rm{e}}^x}{2x} \, \, \stackrel{\text{l'H}}{=} \, \, \lim_{x \to \infty} \, \frac{{\rm{e}}^x}{2} \, \ = \, \, \infty \, . \]

Calculate limit for \(\ln(y) \, \, = \, \, \ln\left( \left( 1 + \sin(4x) \right)^{\cot(x)} \right) \, \,\) and recall, that \(y = {\rm{e}}^{\ln(y)}\).

\(\lim_{x \to 0^+} \, \left( 1 + \sin(4x) \right)^{\cot(x)} = e^4\).

First notice that as \(x \to 0^+\), we have \(1+ \sin(4x) \to 1\) and \(\cot(x) \to \infty\), so the given limit is indeterminate (type \(1^{\infty}\)).

Let \(y = \left( 1 + \sin(4x) \right)^{\cot(x)}\). Then \[ \ln(y) \, \, = \, \, \ln\left( \left( 1 + \sin(4x) \right)^{\cot(x)} \right) \, \, = \, \, \cot(x) \cdot \ln\left( 1 + \sin(4x) \right) \, \, = \, \, \frac{\ln\left( 1 + \sin(4x) \right)}{\tan(x)} \] so l'Hospital's Rule (\(\text{l'H}\)) gives \[ \lim_{x \to 0^+} \, \ln(y) \, \, = \, \, \lim_{x \to 0^+} \, \frac{\ln\left( 1 + \sin(4x) \right)}{\tan(x)} \, \, \stackrel{\text{l'H}}{=} \, \, \lim_{x \to 0^+} \, \frac{\, \, \frac{4 \, \cos(4x)}{1 + \sin(4x)} \, \,}{\sec^2(x)} \, \, = \, \, 4 \] Hence, \(\lim_{x \to 0^+} \, \left( 1 + \sin(4x) \right)^{\cot(x)} = \lim_{x \to 0^+} \, y = \lim_{x \to 0^+} \, {\rm{e}}^{\ln(y)} = {\rm{e}}^4\).

Note, that \(\Delta y = f(x_2) - f(x_1)\) and \( \textrm{d} y \, \, = \, \, f'(x) \, \textrm{d} x \, \,. \)

\({\bf{a)}}\) \(\Delta y = 0.717625 \), \(\textrm{d} y = 0.7\)
\({\bf{b)}}\) \(\Delta y = 0.140701 \), \(\textrm{d} y = 0.14\)

We have
\(\begin{eqnarray*} f(2) & = & 2^3 + 2^2 - 2 \cdot 2 + 1 \, \, = \, \, 9 \\[1mm] f(2.05) & = & (2.05)^3 + (2.05)^2 - 2 \cdot 2.05 + 1 \, \, = \, \, 9.717625 \\[1mm] \Delta y & = & f(2.05) - f(2) \, \, = \, \, 0.717625 \end{eqnarray*}\)

In general, \[ \textrm{d} y \, \, = \, \, f'(x) \, \textrm{d} x \, \, = \, \, (3 x^2 + 2x - 2) \textrm{d} x \, . \] When \(x = 2\) and \(\textrm{d} x = \Delta x = 0.05\), this becomes

\(\begin{eqnarray*} \textrm{d} y & = & (3 \cdot 2^2 + 2 \cdot 2 - 2) \cdot 0.05 \, \, = \, \, 0.7 \\[1mm] f(2.01) & = & (2.01)^3 + (2.01)^2 - 2 \cdot 2.01 + 1 \, \, = \, \, 9.140701 \\[1mm] \Delta y & = & f(2.01) - f(2) \, \, = \, \, 0.140701 \end{eqnarray*}\)

When \(\textrm{d} x = \Delta x = 0.01\), \[ \textrm{d} y \, \, = \, \, (3 \cdot 2^2 + 2 \cdot 2 - 2) \cdot 0.01 \, \, = \, \, 0.14 \, . \]