Using the power rule: \(f'(x) = 2x\)

At \(x = 3\): \(f'(3) = 2 \cdot 3 = 6\)

This means the slope of the tangent line to the curve \(y = x^2\) at the point \((3, 9)\) is 6.

1. \(f'(x) = 12x^3 - 4x\)

   Using the power rule for each term

2. \(g'(x) = 2\cos(2x)\cos(x) - \sin(2x)\sin(x)\)

   Using the product rule: \(g'(x) = f'(x)g(x) + f(x)g'(x)\)

3. \(h'(x) = \frac{2x}{x^2 + 1}\)

   Using the chain rule: \(h'(x) = \frac{1}{x^2 + 1} \cdot 2x\)

\(\frac{\partial f}{\partial x} = 2xy\) (treating \(y\) as a constant)

\(\frac{\partial f}{\partial y} = x^2 + 3y^2\) (treating \(x\) as a constant)

Using the power rule with \(n = 5\):

\(f'(x) = 5 \cdot x^{5-1} = 5x^4\)

Let \(u(x) = x^2\) and \(v(x) = \sin(x)\)

Then \(u'(x) = 2x\) and \(v'(x) = \cos(x)\)

Using the product rule:

\(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\)

\(f'(x) = 2x \cdot \sin(x) + x^2 \cdot \cos(x)\)

Let \(u(x) = x^2\) and \(v(x) = \cos(x)\)

Then \(u'(x) = 2x\) and \(v'(x) = -\sin(x)\)

Using the quotient rule:

\(f'(x) = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{[v(x)]^2}\)

\(f'(x) = \frac{2x \cdot \cos(x) - x^2 \cdot (-\sin(x))}{[\cos(x)]^2}\)

\(f'(x) = \frac{2x \cdot \cos(x) + x^2 \cdot \sin(x)}{[\cos(x)]^2}\)

Let \(g(x) = x^2\) and \(f(u) = \sin(u)\), so \(f(x) = f(g(x)) = \sin(g(x)) = \sin(x^2)\)

Then \(g'(x) = 2x\) and \(f'(u) = \cos(u)\)

Using the chain rule:

\(f'(x) = f'(g(x)) \cdot g'(x)\)

\(f'(x) = \cos(x^2) \cdot 2x = 2x\cos(x^2)\)

Differentiate both sides with respect to \(x\):

\(\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)\)

\(2x + 2y \cdot \frac{dy}{dx} = 0\)

Solve for \(\frac{dy}{dx}\):

\(2y \cdot \frac{dy}{dx} = -2x\)

\(\frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{y}\)

Let \(y = x^x\)

Take the natural logarithm of both sides:

\(\ln(y) = \ln(x^x) = x\ln(x)\)

Differentiate both sides with respect to \(x\):

\(\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1\)

Solve for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = y \cdot (\ln(x) + 1) = x^x \cdot (\ln(x) + 1)\)

First, find \(f(2)\):

\(f(2) = 2^3 - 2(2) + 1 = 8 - 4 + 1 = 5\)

The point is \((2, 5)\)

Next, find \(f'(x)\):

\(f'(x) = 3x^2 - 2\)

Evaluate at \(x = 2\):

\(f'(2) = 3(2)^2 - 2 = 3(4) - 2 = 12 - 2 = 10\)

The slope of the tangent line is 10

Equation of the tangent line:

\(y - 5 = 10(x - 2)\)

\(y - 5 = 10x - 20\)

\(y = 10x - 15\)

The slope of the normal line is \(-\frac{1}{10}\)

Equation of the normal line:

\(y - 5 = -\frac{1}{10}(x - 2)\)

\(y - 5 = -\frac{1}{10}x + \frac{1}{5}\)

\(y = -\frac{1}{10}x + \frac{1}{5} + 5\)

\(y = -\frac{1}{10}x + \frac{51}{10}\)

Let \(l\) be the length and \(w\) be the width of the rectangle.

The perimeter constraint gives us: \(2l + 2w = 100\), so \(l + w = 50\) or \(l = 50 - w\)

Area function: \(A = l \cdot w = (50 - w) \cdot w = 50w - w^2\)

To maximize, we find \(\frac{dA}{dw}\) and set it to zero:

\(\frac{dA}{dw} = 50 - 2w\)

\(50 - 2w = 0\)

\(w = 25\)

Substituting back: \(l = 50 - w = 50 - 25 = 25\)

Therefore, the rectangle with maximum area has dimensions 25 × 25, making it a square.

The volume of a cone is \(V = \frac{1}{3}\pi r^2 h\)

Due to similar triangles, when the water is at height \(h\), the radius at the surface is \(r = \frac{6}{12}h = \frac{h}{2}\)

So the volume is: \(V = \frac{1}{3}\pi (\frac{h}{2})^2 h = \frac{\pi h^3}{12}\)

We know \(\frac{dV}{dt} = 10\) and want to find \(\frac{dh}{dt}\) when \(h = 4\)

Differentiate the volume equation with respect to time:

\(\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}\)

Substitute the known values:

\(10 = \frac{\pi \cdot 4^2}{4} \cdot \frac{dh}{dt}\)

\(10 = 4\pi \cdot \frac{dh}{dt}\)

\(\frac{dh}{dt} = \frac{10}{4\pi} = \frac{5}{2\pi} \approx 0.796\) meters per minute

As \(x \to 0\), we get \(\frac{\sin(0)}{0} = \frac{0}{0}\), which is an indeterminate form.

Apply L'Hôpital's rule:

\(\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}[\sin(x)]}{\frac{d}{dx}[x]} = \lim_{x \to 0} \frac{\cos(x)}{1} = \cos(0) = 1\)