BC Calculus Extra Practice in Differentiation
- Class: 5H: BC Calculus
- Author: Peter Atlas
- Text: Calculus Finney, Demana, Waits, Kennedy
- Find \(y'\) given \(y=\sec{x^2}.\)
Solution
\(y' = \sec{ \left( x^2 \right) } \tan{ \left( x^2 \right)} 2x \)
- Find \(\frac{dy}{dt}\) given \(y=\cot{u}\) where \(u\) is a differentiable function of \(t\).
Solution
\(\frac{dy}{dt} = -\csc^2{u}\frac{du}{dt}\).
- Find \(y'\) given \(y=\sin^2 {x} + \cos^2 {x}\).
Solution
Don't do too much work here! \(y = 1\). The derivative is \(y = 0\).
- Find \(y'\) given \(y=\sin^{-1}{(3x)}\).
Solution
\(y' = \frac{3}{\sqrt{1 - 9x^2}}\).
- Given \(f\) and \(g\) are differentiable functions, find \(\frac{d}{dx} g(f(x))\).
Solution
\(g'(f(x))f'(x)\).
- Given \(f\) and \(g\) are differentiable functions, find \(\frac{d}{dx} f \left( g(x^2) \right) \).
Solution
\(f' \left( g(x^2) \right) g'(x^2)2x.\)
- Given \(y = f(t)\) and \(x = g(t)\), find \(\frac{dy}{dx}\) in differential form.
Solution
\(\frac{dy}{dx} = \frac{df}{dt} \div \frac{dg}{dt}.\)
- Find the derivative, with respect to \(x\), of \(\csc{(\sin{x})}\)
Solution
\( -\csc{( \sin{x})} cot{(\sin{x})}\cos{x}\)
- Find the derivative of \(\sin{x}\) with respect to \(x^2\) . That is, given \(y = \sin {x}\), find \(\frac{dy}{d(x^2 )}\).
Solution
This would be \( \frac{d (\sin{x})}{dx} \div \frac{d(x^2)}{dx}\text{, or }\frac{\cos{x}}{2x} \).
- Find \( \frac{dy}{dx}\) given \(y + xy = 2\).
Solution
\(y' + xy' + y = 0\text{; }y' = -\frac{y}{1 + x}\).
- Find the equation of the tangent line to \(y = 3x^4\text{ at }x = 1\).
Solution
\(y' = 12x^3\text{. }y'(1) = 12\text{. }y(1) = 3\text{. } y - 3 = 12 (x - 1)\)
- A particle moves along the x-axis according to the equation \(x(t) = t^3- 9t\) where \(t > 0\). Find the value of \(t\) where the particle changes direction.
Solution
\( v(t) = 3t^2 - 9 = 3(t + \sqrt{3})(t - \sqrt{3}). \) Particle changes direction where v changes sign at \(\sqrt{3}\).
- Find the x-intercept of the normal to \(y = 2x^2 - 5x + 2\text{ at }x = 0\).
Solution
\( y' = 2x - 5\text{. }y'(0) = -5\text{. } y(0) = 2\text{. Normal: } y - 2 = \frac{1}{5}(x - 0)\). The x-intercept is at -10.
- The cost of slicing \(c\) cows is given by \(S(c) = 200(c - 4)^3 + 400(3c - 4)^2 \), where \(S\) is measured in dollars, and \(c\) is measured in cows. Find the marginal cost of slicing the fifth cow.
Solution
\(S'(c) = 600(c - 4)^2 + 800(3c - 4)(3)\text{. } S'(4) = 600(4 - 4)^2 + 800(3 \cdot 4 - 4)(3) = 19200 \)
- Find the derivative of \(y = (3x + 2)^3\) evaluated at \(x = 1\).
Solution
\(y' = 3(3x+2)^2 \cdot 3. y'(1) = 3 \cdot 5^2 \cdot 3\).
- Find the derivative of \(y = \sin{x} \cos{x}\) evaluated at \(x = 0\).
Solution
\(y = \frac{1}{2} \sin{(2x)}\text{, so }y' = \frac{1}{2} \cos{(2x)} \cdot 2 = \cos{(2x)}\text{. }y'(0) = 1\).
- Find the first positive value of \(t\) at which the curve
\( \displaystyle \begin{align*}
y &= \sin { \left( t + \frac{ \pi}{ 4} \right) } \\
x &= \tan{t} \end{align*} \)
has a horizontal tangent.
Solution
\( \frac{dy}{dt} = \cos{(t + \frac{\pi}{4} )}\). \( \frac{dx}{dt} = \sec^2 {t}\). \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{\cos{(t + \frac{\pi}{4} )}}{ \sec^2 {t}}\). This will have a horizontal tangent where the numerator is zero and the denominator isn't. \(\cos {(t + \frac{\pi}{4} )} = 0 \implies t = \frac{\pi}{4}\) .
- Given \(y = \cos{t}\), \(x = \sin{t}\), find \(\frac{d^2 y}{dx^2} \mid_{t = \frac{\pi}{3}}\).
Solution
\(\frac{dy}{dt} = -\sin{t}\text{. }\frac{dx}{dt} = \cos{t}\text{. }\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = -\tan{t}\). \(y'' = \frac{dy'}{dt} \div \frac{dx}{dt} = -\frac{\sec^2 {t}}{\cos{t}} = -\sec^3{t}\). At \(t = \frac{\pi}{3}\), the second derivative is \(-\sec^3 {\frac{\pi}{3}}.\)
- Given \(f\) is a continuous, differentiable function where \(f(0) = 3\), \(f(1) = 2\), \(f(2) = 3\), \(f'(0) = 6\), \(f'(1) = 7\), and \(f'(2) = 8\), find \(\displaystyle \lim_{x\to 1}\frac{f(x) - 2}{x - 1}\).
Solution
This is the definition of \(f'(x)\) evaluated at \(x = 1\). The answer is \(f'(1) = 7\).
- Find the derivative of \(y = |x|\) at \(x = 0\). If it doesn't exist, so state and explain why.
Solution
D.N.E. as the left- and right-handed limits of the difference quotient disagree.
- Given a continuous curve that has roots at 0, 3, 5, and 9, in at least how many places must the derivative of the curve be 0?
Solution
None. It never says the function must be differentiable.