If \( y = (x^2 + 1) \tan x \), then \( \frac{dy}{dx} = \)
Applying the product rule to the function:
\[ y = (x^2 + 1) \tan x \]
\[ \frac{dy}{dx} = (x^2 + 1) \sec^2 x + \tan x (2x) \]
If \( h(x) = \begin{cases} x^2 + a, & \text{for } x < -1 \\ x^3 - 8, & \text{for } x \geq -1 \end{cases} \) determine all values of \( a \) so that \( h \) is continuous for all values of \( x \).
Continuous at \( x = b \) implies:
\( x^2 + a \) and \( x^3 - 8 \) are polynomials, so they are defined and continuous on their own subdomains.
The only place to check is at \( x = -1 \).
Is \( h(-1) \) defined? Yes. \( h(x) = x^3 - 8 \) for \( x \geq -1 \), so \( h(-1) = (-1)^3 - 8 = -9 \).
Does \( \lim_{x \to -1} h(x) \) exist? Check if \( \lim_{x \to -1^-} h(x) = \lim_{x \to -1^+} h(x) \).
\[ \lim_{x \to -1^-} h(x) = \lim_{x \to -1^-} x^2 + a = 1 + a \]
\[ \lim_{x \to -1^+} h(x) = \lim_{x \to -1^+} x^3 - 8 = -9 \]
They match if \( 1 + a = -9 \) or \( a = -10 \).
Lastly, is \( h(-1) = \lim_{x \to -1} h(x) \) with \( a = -10 \)?
\( h(-1) = -9 \)
\( \lim_{x \to -1} h(x) = 1 + a = 1 - 10 = -9 \)
so, yes.
Evaluate \( \lim_{x \to 0^+} x \cos(\frac{1}{x}) \). (Hint: \( -1 \le \cos(\frac{1}{x}) \le 1 \) for all \( x \neq 0 \).)
if \( g(x) \le f(x) \le h(x) \) for all \( x \) except possibly at \( x = a \)
then \( \lim_{x \to a} g(x) \le \lim_{x \to a} f(x) \le \lim_{x \to a} h(x) \)
\( -1 \le \cos(\frac{1}{x}) \le 1 \)
multiply by \( x \)
\( -x \le x \cos(\frac{1}{x}) \le x \)
\( \lim_{x \to 0^+} -x \le \lim_{x \to 0^+} x \cos(\frac{1}{x}) \le \lim_{x \to 0^+} x \)
\( 0 \le \lim_{x \to 0^+} x \cos(\frac{1}{x}) \le 0 \)
so limit is 0
If \( f(x) = \frac{1}{x+3} \), then \( \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} = \)
\( f(x) = \frac{1}{x+3} \)
\( f(1) = \frac{1}{4} \)
Recognize that \( f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \). Here, \( a = 1 \).
So just calculate \( f'(1) \):
If \( f(x) = \frac{1-x}{1+x} \), then \( f'(1) = \)
Evaluating at \( x = 1 \):
Find \( f''(x) \) if \( f(x) = \frac{1-x}{1+x} \)
Simplify:
Note: The following calculation is crossed out in the original notes as an inefficient method.
Chain Rule applied to denominator:
Rewriting the first derivative to use the power rule instead of the quotient rule:
Applying the power rule and chain rule to find the second derivative:
The correct answer is A.
If \( y = \ln(1 - x^2) + \sin^2 x \), then \( \frac{dy}{dx} = \)
\( y = \ln(1 - x^2) + (\sin x)^2 \)
\( \frac{dy}{dx} = \frac{1}{1-x^2} \frac{d}{dx}(1-x^2) + 2(\sin x) \frac{d}{dx}(\sin x) \)
\( = \frac{-2x}{1-x^2} + 2 \sin x \cos x \)
\( \frac{d}{dx} \ln(u) = \frac{1}{u} \frac{du}{dx} \)
\( \frac{d}{dx} u^n = n u^{n-1} \frac{du}{dx} \)
Assume that \( y \) is defined implicitly as a differentiable function of \( x \) by the equation \( xy^2 - x^2 + y + 5 = 0 \). Find \( \frac{dy}{dx} \) at \( (-2, 1) \).
\( \frac{d}{dx}(xy^2 - x^2 + y + 5) = \frac{d}{dx}(0) \)
Applying the product rule to \( xy^2 \):
\( (x)(2)(y)\frac{dy}{dx} + (y^2)(1) - 2x + \frac{dy}{dx} + 0 = 0 \)
\( 2xy \frac{dy}{dx} + \frac{dy}{dx} = 2x - y^2 \)
\( \frac{dy}{dx}(2xy + 1) = 2x - y^2 \)
\( \frac{dy}{dx} = \frac{2x - y^2}{2xy + 1} \)
Now we have \( \frac{dy}{dx} \), we can use \( (x, y) = (-2, 1) \)
\( = \frac{2(-2) - (1)^2}{2(-2)(1) + 1} = \frac{-5}{-3} = \frac{5}{3} \)
Page 11
Find the maximum and minimum values of the function \( f(x) = 3x^2 + 6x - 10 \) on the interval \( -2 \le x \le 2 \).
Find where \( f'(x) = 0 \) or \( f' \) DNE (Does Not Exist), then compare \( f(x) \) at those points and the end points.
\( f(x) = 3x^2 + 6x - 10 \)
\( f'(x) = 6x + 6 = 0 \implies x = -1 \)
Check if this is inside the interval. Here, we keep it.
Now compare (using the original function, NOT \( f' \)):
Page 12
For a differentiable function \( f(x) \) it is known that \( f(3) = 5 \) and \( f'(3) = -2 \). Use a linear approximation to get the approximate value of \( f(3.02) \).
\( f(x) \approx f(a) + f'(a)(x - a) \)
Here, we have \( f(3) \) and \( f'(3) \) so we use \( a = 3 \).
\( f(x) \approx f(3) + f'(3)(x - 3) \)
\( f(x) \approx 5 - 2(x - 3) \)
\( f(3.02) \approx 5 - 2(3.02 - 3) = 5 - 2(0.02) = 5 - 0.04 = 4.96 \)
Water is withdrawn from a conical reservoir, 8 feet in diameter and 10 feet deep (vertex down) at the constant rate of 5 ft3/min. How fast is the water level falling when the depth of the water in the reservoir is 5 ft? (\[V = \frac{1}{3}\pi r^2 h\]).
Water goes out at 5 ft3/min
we know and want to know more about \(h\), so we want to get rid of \(r\)
they are similar (same shape)
Sub into V equation
Substituting known values: \(\frac{dv}{dt} = -5\) and \(h = 5\)
so water level is falling at \(\frac{5}{4\pi}\) ft/min